WORST_CASE(Omega(n^1),O(n^1))
proof of /export/starexec/sandbox/benchmark/theBenchmark.trs
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).

(0) CpxTRS
(1) CpxTrsToCdtProof [UPPER BOUND(ID), 0 ms]
(2) CdtProblem
    (3) CdtLeafRemovalProof [BOTH BOUNDS(ID, ID), 0 ms]
    (4) CdtProblem
    (5) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 133 ms]
    (6) CdtProblem
    (7) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 33 ms]
    (10) CdtProblem
    (11) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (12) CdtProblem
    (13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 26 ms]
    (14) CdtProblem
    (15) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (16) CdtProblem
    (17) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 27 ms]
    (18) CdtProblem
    (19) CdtKnowledgeProof [FINISHED, 0 ms]
    (20) BOUNDS(1, 1)
(21) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
(22) CpxTRS
    (23) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (24) typed CpxTrs
    (25) OrderProof [LOWER BOUND(ID), 0 ms]
    (26) typed CpxTrs
    (27) RewriteLemmaProof [LOWER BOUND(ID), 271 ms]
    (28) BEST
        (29) proven lower bound
            (30) LowerBoundPropagationProof [FINISHED, 0 ms]
            (31) BOUNDS(n^1, INF)
        (32) typed CpxTrs
            (33) RewriteLemmaProof [LOWER BOUND(ID), 0 ms]
            (34) typed CpxTrs
            (35) RewriteLemmaProof [LOWER BOUND(ID), 87 ms]
            (36) typed CpxTrs
            (37) RewriteLemmaProof [LOWER BOUND(ID), 36 ms]
            (38) typed CpxTrs


----------------------------------------

(0)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, n^1).


The TRS R consists of the following rules:

   append(@l1, @l2) -> append#1(@l1, @l2)
   append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
   append#1(nil, @l2) -> @l2
   appendAll(@l) -> appendAll#1(@l)
   appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
   appendAll#1(nil) -> nil
   appendAll2(@l) -> appendAll2#1(@l)
   appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
   appendAll2#1(nil) -> nil
   appendAll3(@l) -> appendAll3#1(@l)
   appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
   appendAll3#1(nil) -> nil

S is empty.
Rewrite Strategy: INNERMOST
----------------------------------------

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPEND#1(nil, z0) -> c2
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL#1(nil) -> c5
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL2#1(nil) -> c8
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL3#1(nil) -> c11
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPEND#1(nil, z0) -> c2
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL#1(nil) -> c5
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL2#1(nil) -> c8
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL3#1(nil) -> c11
K tuples:none
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c2, c3_1, c4_2, c5, c6_1, c7_3, c8, c9_1, c10_3, c11


----------------------------------------

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 4 trailing nodes:
   APPENDALL#1(nil) -> c5
   APPENDALL2#1(nil) -> c8
   APPEND#1(nil, z0) -> c2
   APPENDALL3#1(nil) -> c11

----------------------------------------

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
K tuples:none
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(5) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
We considered the (Usable) Rules:none
And the Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(::(x_1, x_2)) = [1] + x_1 + x_2
   POL(APPEND(x_1, x_2)) = 0
   POL(APPEND#1(x_1, x_2)) = 0
   POL(APPENDALL(x_1)) = 0
   POL(APPENDALL#1(x_1)) = 0
   POL(APPENDALL2(x_1)) = 0
   POL(APPENDALL2#1(x_1)) = 0
   POL(APPENDALL3(x_1)) = [1] + x_1
   POL(APPENDALL3#1(x_1)) = x_1
   POL(append(x_1, x_2)) = 0
   POL(append#1(x_1, x_2)) = [1] + x_1 + x_2
   POL(appendAll(x_1)) = x_1
   POL(appendAll#1(x_1)) = [1] + x_1
   POL(appendAll2(x_1)) = x_1
   POL(appendAll2#1(x_1)) = [1] + x_1
   POL(appendAll3(x_1)) = [1] + x_1
   POL(appendAll3#1(x_1)) = [1] + x_1
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c10(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c3(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c9(x_1)) = x_1
   POL(nil) = [1]

----------------------------------------

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(7) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))

----------------------------------------

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
We considered the (Usable) Rules:none
And the Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(::(x_1, x_2)) = [1] + x_1 + x_2
   POL(APPEND(x_1, x_2)) = 0
   POL(APPEND#1(x_1, x_2)) = 0
   POL(APPENDALL(x_1)) = 0
   POL(APPENDALL#1(x_1)) = 0
   POL(APPENDALL2(x_1)) = x_1
   POL(APPENDALL2#1(x_1)) = x_1
   POL(APPENDALL3(x_1)) = [1] + x_1
   POL(APPENDALL3#1(x_1)) = [1] + x_1
   POL(append(x_1, x_2)) = 0
   POL(append#1(x_1, x_2)) = [1] + x_1 + x_2
   POL(appendAll(x_1)) = x_1
   POL(appendAll#1(x_1)) = [1] + x_1
   POL(appendAll2(x_1)) = x_1
   POL(appendAll2#1(x_1)) = [1] + x_1
   POL(appendAll3(x_1)) = [1] + x_1
   POL(appendAll3#1(x_1)) = [1] + x_1
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c10(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c3(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c9(x_1)) = x_1
   POL(nil) = [1]

----------------------------------------

(10)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(11) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))

----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
We considered the (Usable) Rules:none
And the Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(::(x_1, x_2)) = [1] + x_1 + x_2
   POL(APPEND(x_1, x_2)) = 0
   POL(APPEND#1(x_1, x_2)) = 0
   POL(APPENDALL(x_1)) = x_1
   POL(APPENDALL#1(x_1)) = x_1
   POL(APPENDALL2(x_1)) = x_1
   POL(APPENDALL2#1(x_1)) = x_1
   POL(APPENDALL3(x_1)) = x_1
   POL(APPENDALL3#1(x_1)) = x_1
   POL(append(x_1, x_2)) = 0
   POL(append#1(x_1, x_2)) = [1] + x_1 + x_2
   POL(appendAll(x_1)) = x_1
   POL(appendAll#1(x_1)) = [1] + x_1
   POL(appendAll2(x_1)) = x_1
   POL(appendAll2#1(x_1)) = [1] + x_1
   POL(appendAll3(x_1)) = [1] + x_1
   POL(appendAll3#1(x_1)) = [1] + x_1
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c10(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c3(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c9(x_1)) = x_1
   POL(nil) = [1]

----------------------------------------

(14)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(15) CdtKnowledgeProof (BOTH BOUNDS(ID, ID))
The following tuples could be moved from S to K by knowledge propagation:
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))

----------------------------------------

(16)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(17) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
We considered the (Usable) Rules:
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   append(z0, z1) -> append#1(z0, z1)
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll#1(nil) -> nil
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
And the Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(::(x_1, x_2)) = [2] + x_1 + x_2
   POL(APPEND(x_1, x_2)) = x_1
   POL(APPEND#1(x_1, x_2)) = x_1
   POL(APPENDALL(x_1)) = x_1
   POL(APPENDALL#1(x_1)) = x_1
   POL(APPENDALL2(x_1)) = [2]x_1
   POL(APPENDALL2#1(x_1)) = [2]x_1
   POL(APPENDALL3(x_1)) = [2] + [3]x_1
   POL(APPENDALL3#1(x_1)) = [2] + [3]x_1
   POL(append(x_1, x_2)) = x_1 + x_2
   POL(append#1(x_1, x_2)) = x_1 + x_2
   POL(appendAll(x_1)) = x_1
   POL(appendAll#1(x_1)) = x_1
   POL(appendAll2(x_1)) = x_1
   POL(appendAll2#1(x_1)) = x_1
   POL(appendAll3(x_1)) = [3] + [3]x_1
   POL(appendAll3#1(x_1)) = [3] + [3]x_1
   POL(c(x_1)) = x_1
   POL(c1(x_1)) = x_1
   POL(c10(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c3(x_1)) = x_1
   POL(c4(x_1, x_2)) = x_1 + x_2
   POL(c6(x_1)) = x_1
   POL(c7(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(c9(x_1)) = x_1
   POL(nil) = 0

----------------------------------------

(18)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   append(z0, z1) -> append#1(z0, z1)
   append#1(::(z0, z1), z2) -> ::(z0, append(z1, z2))
   append#1(nil, z0) -> z0
   appendAll(z0) -> appendAll#1(z0)
   appendAll#1(::(z0, z1)) -> append(z0, appendAll(z1))
   appendAll#1(nil) -> nil
   appendAll2(z0) -> appendAll2#1(z0)
   appendAll2#1(::(z0, z1)) -> append(appendAll(z0), appendAll2(z1))
   appendAll2#1(nil) -> nil
   appendAll3(z0) -> appendAll3#1(z0)
   appendAll3#1(::(z0, z1)) -> append(appendAll2(z0), appendAll3(z1))
   appendAll3#1(nil) -> nil
Tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
S tuples:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
K tuples:
   APPENDALL3(z0) -> c9(APPENDALL3#1(z0))
   APPENDALL3#1(::(z0, z1)) -> c10(APPEND(appendAll2(z0), appendAll3(z1)), APPENDALL2(z0), APPENDALL3(z1))
   APPENDALL2#1(::(z0, z1)) -> c7(APPEND(appendAll(z0), appendAll2(z1)), APPENDALL(z0), APPENDALL2(z1))
   APPENDALL2(z0) -> c6(APPENDALL2#1(z0))
   APPENDALL#1(::(z0, z1)) -> c4(APPEND(z0, appendAll(z1)), APPENDALL(z1))
   APPENDALL(z0) -> c3(APPENDALL#1(z0))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
Defined Rule Symbols:   append_2, append#1_2, appendAll_1, appendAll#1_1, appendAll2_1, appendAll2#1_1, appendAll3_1, appendAll3#1_1

Defined Pair Symbols:   APPEND_2, APPEND#1_2, APPENDALL_1, APPENDALL#1_1, APPENDALL2_1, APPENDALL2#1_1, APPENDALL3_1, APPENDALL3#1_1

Compound Symbols:   c_1, c1_1, c3_1, c4_2, c6_1, c7_3, c9_1, c10_3


----------------------------------------

(19) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   APPEND(z0, z1) -> c(APPEND#1(z0, z1))
   APPEND#1(::(z0, z1), z2) -> c1(APPEND(z1, z2))
Now S is empty
----------------------------------------

(20)
BOUNDS(1, 1)

----------------------------------------

(21) RenamingProof (BOTH BOUNDS(ID, ID))
Renamed function symbols to avoid clashes with predefined symbol.
----------------------------------------

(22)
Obligation:
The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   append(@l1, @l2) -> append#1(@l1, @l2)
   append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
   append#1(nil, @l2) -> @l2
   appendAll(@l) -> appendAll#1(@l)
   appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
   appendAll#1(nil) -> nil
   appendAll2(@l) -> appendAll2#1(@l)
   appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
   appendAll2#1(nil) -> nil
   appendAll3(@l) -> appendAll3#1(@l)
   appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
   appendAll3#1(nil) -> nil

S is empty.
Rewrite Strategy: INNERMOST
----------------------------------------

(23) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Infered types.
----------------------------------------

(24)
Obligation:
Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil

----------------------------------------

(25) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
append, append#1, appendAll, appendAll#1, appendAll2, appendAll2#1, appendAll3, appendAll3#1

They will be analysed ascendingly in the following order:
append = append#1
append < appendAll#1
append < appendAll2#1
append < appendAll3#1
appendAll = appendAll#1
appendAll < appendAll2#1
appendAll2 = appendAll2#1
appendAll2 < appendAll3#1
appendAll3 = appendAll3#1

----------------------------------------

(26)
Obligation:
Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil


Generator Equations:
gen_:::nil2_0(0) <=> nil
gen_:::nil2_0(+(x, 1)) <=> ::(nil, gen_:::nil2_0(x))


The following defined symbols remain to be analysed:
append#1, append, appendAll, appendAll#1, appendAll2, appendAll2#1, appendAll3, appendAll3#1

They will be analysed ascendingly in the following order:
append = append#1
append < appendAll#1
append < appendAll2#1
append < appendAll3#1
appendAll = appendAll#1
appendAll < appendAll2#1
appendAll2 = appendAll2#1
appendAll2 < appendAll3#1
appendAll3 = appendAll3#1

----------------------------------------

(27) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
append#1(gen_:::nil2_0(n4_0), gen_:::nil2_0(b)) -> gen_:::nil2_0(+(n4_0, b)), rt in Omega(1 + n4_0)

Induction Base:
append#1(gen_:::nil2_0(0), gen_:::nil2_0(b)) ->_R^Omega(1)
gen_:::nil2_0(b)

Induction Step:
append#1(gen_:::nil2_0(+(n4_0, 1)), gen_:::nil2_0(b)) ->_R^Omega(1)
::(nil, append(gen_:::nil2_0(n4_0), gen_:::nil2_0(b))) ->_R^Omega(1)
::(nil, append#1(gen_:::nil2_0(n4_0), gen_:::nil2_0(b))) ->_IH
::(nil, gen_:::nil2_0(+(b, c5_0)))

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(28)
Complex Obligation (BEST)

----------------------------------------

(29)
Obligation:
Proved the lower bound n^1 for the following obligation:

Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil


Generator Equations:
gen_:::nil2_0(0) <=> nil
gen_:::nil2_0(+(x, 1)) <=> ::(nil, gen_:::nil2_0(x))


The following defined symbols remain to be analysed:
append#1, append, appendAll, appendAll#1, appendAll2, appendAll2#1, appendAll3, appendAll3#1

They will be analysed ascendingly in the following order:
append = append#1
append < appendAll#1
append < appendAll2#1
append < appendAll3#1
appendAll = appendAll#1
appendAll < appendAll2#1
appendAll2 = appendAll2#1
appendAll2 < appendAll3#1
appendAll3 = appendAll3#1

----------------------------------------

(30) LowerBoundPropagationProof (FINISHED)
Propagated lower bound.
----------------------------------------

(31)
BOUNDS(n^1, INF)

----------------------------------------

(32)
Obligation:
Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil


Lemmas:
append#1(gen_:::nil2_0(n4_0), gen_:::nil2_0(b)) -> gen_:::nil2_0(+(n4_0, b)), rt in Omega(1 + n4_0)


Generator Equations:
gen_:::nil2_0(0) <=> nil
gen_:::nil2_0(+(x, 1)) <=> ::(nil, gen_:::nil2_0(x))


The following defined symbols remain to be analysed:
append, appendAll, appendAll#1, appendAll2, appendAll2#1, appendAll3, appendAll3#1

They will be analysed ascendingly in the following order:
append = append#1
append < appendAll#1
append < appendAll2#1
append < appendAll3#1
appendAll = appendAll#1
appendAll < appendAll2#1
appendAll2 = appendAll2#1
appendAll2 < appendAll3#1
appendAll3 = appendAll3#1

----------------------------------------

(33) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
appendAll#1(gen_:::nil2_0(n633_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n633_0)

Induction Base:
appendAll#1(gen_:::nil2_0(0)) ->_R^Omega(1)
nil

Induction Step:
appendAll#1(gen_:::nil2_0(+(n633_0, 1))) ->_R^Omega(1)
append(nil, appendAll(gen_:::nil2_0(n633_0))) ->_R^Omega(1)
append(nil, appendAll#1(gen_:::nil2_0(n633_0))) ->_IH
append(nil, gen_:::nil2_0(0)) ->_R^Omega(1)
append#1(nil, gen_:::nil2_0(0)) ->_L^Omega(1)
gen_:::nil2_0(+(0, 0))

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(34)
Obligation:
Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil


Lemmas:
append#1(gen_:::nil2_0(n4_0), gen_:::nil2_0(b)) -> gen_:::nil2_0(+(n4_0, b)), rt in Omega(1 + n4_0)
appendAll#1(gen_:::nil2_0(n633_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n633_0)


Generator Equations:
gen_:::nil2_0(0) <=> nil
gen_:::nil2_0(+(x, 1)) <=> ::(nil, gen_:::nil2_0(x))


The following defined symbols remain to be analysed:
appendAll, appendAll2, appendAll2#1, appendAll3, appendAll3#1

They will be analysed ascendingly in the following order:
appendAll = appendAll#1
appendAll < appendAll2#1
appendAll2 = appendAll2#1
appendAll2 < appendAll3#1
appendAll3 = appendAll3#1

----------------------------------------

(35) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
appendAll2#1(gen_:::nil2_0(n1017_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n1017_0)

Induction Base:
appendAll2#1(gen_:::nil2_0(0)) ->_R^Omega(1)
nil

Induction Step:
appendAll2#1(gen_:::nil2_0(+(n1017_0, 1))) ->_R^Omega(1)
append(appendAll(nil), appendAll2(gen_:::nil2_0(n1017_0))) ->_R^Omega(1)
append(appendAll#1(nil), appendAll2(gen_:::nil2_0(n1017_0))) ->_L^Omega(1)
append(gen_:::nil2_0(0), appendAll2(gen_:::nil2_0(n1017_0))) ->_R^Omega(1)
append(gen_:::nil2_0(0), appendAll2#1(gen_:::nil2_0(n1017_0))) ->_IH
append(gen_:::nil2_0(0), gen_:::nil2_0(0)) ->_R^Omega(1)
append#1(gen_:::nil2_0(0), gen_:::nil2_0(0)) ->_L^Omega(1)
gen_:::nil2_0(+(0, 0))

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(36)
Obligation:
Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil


Lemmas:
append#1(gen_:::nil2_0(n4_0), gen_:::nil2_0(b)) -> gen_:::nil2_0(+(n4_0, b)), rt in Omega(1 + n4_0)
appendAll#1(gen_:::nil2_0(n633_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n633_0)
appendAll2#1(gen_:::nil2_0(n1017_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n1017_0)


Generator Equations:
gen_:::nil2_0(0) <=> nil
gen_:::nil2_0(+(x, 1)) <=> ::(nil, gen_:::nil2_0(x))


The following defined symbols remain to be analysed:
appendAll2, appendAll3, appendAll3#1

They will be analysed ascendingly in the following order:
appendAll2 = appendAll2#1
appendAll2 < appendAll3#1
appendAll3 = appendAll3#1

----------------------------------------

(37) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
appendAll3#1(gen_:::nil2_0(n1767_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n1767_0)

Induction Base:
appendAll3#1(gen_:::nil2_0(0)) ->_R^Omega(1)
nil

Induction Step:
appendAll3#1(gen_:::nil2_0(+(n1767_0, 1))) ->_R^Omega(1)
append(appendAll2(nil), appendAll3(gen_:::nil2_0(n1767_0))) ->_R^Omega(1)
append(appendAll2#1(nil), appendAll3(gen_:::nil2_0(n1767_0))) ->_L^Omega(1)
append(gen_:::nil2_0(0), appendAll3(gen_:::nil2_0(n1767_0))) ->_R^Omega(1)
append(gen_:::nil2_0(0), appendAll3#1(gen_:::nil2_0(n1767_0))) ->_IH
append(gen_:::nil2_0(0), gen_:::nil2_0(0)) ->_R^Omega(1)
append#1(gen_:::nil2_0(0), gen_:::nil2_0(0)) ->_L^Omega(1)
gen_:::nil2_0(+(0, 0))

We have rt in Omega(n^1) and sz in O(n). Thus, we have irc_R in Omega(n).
----------------------------------------

(38)
Obligation:
Innermost TRS:
Rules:
append(@l1, @l2) -> append#1(@l1, @l2)
append#1(::(@x, @xs), @l2) -> ::(@x, append(@xs, @l2))
append#1(nil, @l2) -> @l2
appendAll(@l) -> appendAll#1(@l)
appendAll#1(::(@l1, @ls)) -> append(@l1, appendAll(@ls))
appendAll#1(nil) -> nil
appendAll2(@l) -> appendAll2#1(@l)
appendAll2#1(::(@l1, @ls)) -> append(appendAll(@l1), appendAll2(@ls))
appendAll2#1(nil) -> nil
appendAll3(@l) -> appendAll3#1(@l)
appendAll3#1(::(@l1, @ls)) -> append(appendAll2(@l1), appendAll3(@ls))
appendAll3#1(nil) -> nil

Types:
append :: :::nil -> :::nil -> :::nil
append#1 :: :::nil -> :::nil -> :::nil
:: :: :::nil -> :::nil -> :::nil
nil :: :::nil
appendAll :: :::nil -> :::nil
appendAll#1 :: :::nil -> :::nil
appendAll2 :: :::nil -> :::nil
appendAll2#1 :: :::nil -> :::nil
appendAll3 :: :::nil -> :::nil
appendAll3#1 :: :::nil -> :::nil
hole_:::nil1_0 :: :::nil
gen_:::nil2_0 :: Nat -> :::nil


Lemmas:
append#1(gen_:::nil2_0(n4_0), gen_:::nil2_0(b)) -> gen_:::nil2_0(+(n4_0, b)), rt in Omega(1 + n4_0)
appendAll#1(gen_:::nil2_0(n633_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n633_0)
appendAll2#1(gen_:::nil2_0(n1017_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n1017_0)
appendAll3#1(gen_:::nil2_0(n1767_0)) -> gen_:::nil2_0(0), rt in Omega(1 + n1767_0)


Generator Equations:
gen_:::nil2_0(0) <=> nil
gen_:::nil2_0(+(x, 1)) <=> ::(nil, gen_:::nil2_0(x))


The following defined symbols remain to be analysed:
appendAll3

They will be analysed ascendingly in the following order:
appendAll3 = appendAll3#1