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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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)), 130 ms]
    (6) CdtProblem
    (7) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (8) CdtProblem
    (9) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 34 ms]
    (10) CdtProblem
    (11) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (12) CdtProblem
    (13) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 33 ms]
    (14) CdtProblem
    (15) CdtKnowledgeProof [BOTH BOUNDS(ID, ID), 0 ms]
    (16) CdtProblem
    (17) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 29 ms]
    (18) CdtProblem
    (19) CdtKnowledgeProof [FINISHED, 0 ms]
    (20) BOUNDS(1, 1)
(21) RelTrsToDecreasingLoopProblemProof [LOWER BOUND(ID), 0 ms]
(22) TRS for Loop Detection
    (23) DecreasingLoopProof [LOWER BOUND(ID), 20 ms]
    (24) BEST
        (25) proven lower bound
            (26) LowerBoundPropagationProof [FINISHED, 0 ms]
            (27) BOUNDS(n^1, INF)
        (28) TRS for Loop Detection


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

(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:
   APPENDALL2#1(nil) -> c8
   APPEND#1(nil, z0) -> c2
   APPENDALL3#1(nil) -> c11
   APPENDALL#1(nil) -> c5

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

(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) RelTrsToDecreasingLoopProblemProof (LOWER BOUND(ID))
Transformed a relative TRS into a decreasing-loop problem.
----------------------------------------

(22)
Obligation:
Analyzing the following TRS for decreasing loops:

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

(23) DecreasingLoopProof (LOWER BOUND(ID))
The following loop(s) give(s) rise to the lower bound Omega(n^1):

The rewrite sequence

appendAll#1(::(@l1, @ls)) ->^+ append(@l1, appendAll#1(@ls))

gives rise to a decreasing loop by considering the right hand sides subterm at position [1].

The pumping substitution is [@ls / ::(@l1, @ls)].

The result substitution is [ ].




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

(24)
Complex Obligation (BEST)

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

(25)
Obligation:
Proved the lower bound n^1 for the following 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
----------------------------------------

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

(27)
BOUNDS(n^1, INF)

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

(28)
Obligation:
Analyzing the following TRS for decreasing loops:

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