WORST_CASE(Omega(n^1),O(n^1))
proof of input_6bYvhddjNF.trs
# AProVE Commit ID: aff8ecad908e01718a4c36e68d2e55d5e0f16e15 fuhs 20220216 unpublished


The Runtime Complexity (parallel-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) CdtUsableRulesProof [BOTH BOUNDS(ID, ID), 0 ms]
    (6) CdtProblem
    (7) CdtRuleRemovalProof [UPPER BOUND(ADD(n^1)), 40 ms]
    (8) CdtProblem
    (9) CdtKnowledgeProof [FINISHED, 0 ms]
    (10) BOUNDS(1, 1)
(11) CpxTrsToCdtProof [BOTH BOUNDS(ID, ID), 0 ms]
(12) CdtProblem
    (13) CdtToCpxRelTrsProof [BOTH BOUNDS(ID, ID), 0 ms]
    (14) CpxRelTRS
    (15) RenamingProof [BOTH BOUNDS(ID, ID), 0 ms]
    (16) CpxRelTRS
    (17) TypeInferenceProof [BOTH BOUNDS(ID, ID), 0 ms]
    (18) typed CpxTrs
    (19) OrderProof [LOWER BOUND(ID), 8 ms]
    (20) typed CpxTrs
    (21) RewriteLemmaProof [LOWER BOUND(ID), 649 ms]
    (22) BEST
        (23) proven lower bound
            (24) LowerBoundPropagationProof [FINISHED, 0 ms]
            (25) BOUNDS(n^1, INF)
        (26) typed CpxTrs
            (27) RewriteLemmaProof [LOWER BOUND(ID), 165 ms]
            (28) BOUNDS(1, INF)


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

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


The TRS R consists of the following rules:

   quot(0, s(y), s(z)) -> 0
   quot(s(x), s(y), z) -> quot(x, y, z)
   quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

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

(1) CpxTrsToCdtProof (UPPER BOUND(ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
----------------------------------------

(2)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))
Tuples:
   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:   quot_3

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c, c1_1, c2_1


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

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID))
Removed 1 trailing nodes:
   QUOT(0, s(z0), s(z1)) -> c

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

(4)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))
Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:   quot_3

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c1_1, c2_1


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

(5) CdtUsableRulesProof (BOTH BOUNDS(ID, ID))
The following rules are not usable and were removed:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))

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

(6)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:none

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c1_1, c2_1


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

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)))
Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
The order we found is given by the following interpretation:

Polynomial interpretation :

   POL(0) = 0
   POL(QUOT(x_1, x_2, x_3)) = x_1
   POL(c1(x_1)) = x_1
   POL(c2(x_1)) = x_1
   POL(s(x_1)) = [1] + x_1

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

(8)
Obligation:
Complexity Dependency Tuples Problem

Rules:none
Tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
Defined Rule Symbols:none

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c1_1, c2_1


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

(9) CdtKnowledgeProof (FINISHED)
The following tuples could be moved from S to K by knowledge propagation:
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
Now S is empty
----------------------------------------

(10)
BOUNDS(1, 1)

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

(11) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID))
Converted Cpx (relative) TRS with rewrite strategy PARALLEL_INNERMOST to CDT
----------------------------------------

(12)
Obligation:
Complexity Dependency Tuples Problem

Rules:
   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))
Tuples:
   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
S tuples:
   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
K tuples:none
Defined Rule Symbols:   quot_3

Defined Pair Symbols:   QUOT_3

Compound Symbols:   c, c1_1, c2_1


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

(13) CdtToCpxRelTrsProof (BOTH BOUNDS(ID, ID))
Converted S to standard rules, and D \ S as well as R to relative rules.
----------------------------------------

(14)
Obligation:
The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   QUOT(0, s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0, s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))

The (relative) TRS S consists of the following rules:

   quot(0, s(z0), s(z1)) -> 0
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0, s(z1)) -> s(quot(z0, s(z1), s(z1)))

Rewrite Strategy: INNERMOST
----------------------------------------

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

(16)
Obligation:
The Runtime Complexity (innermost) of the given CpxRelTRS could be proven to be BOUNDS(n^1, INF).


The TRS R consists of the following rules:

   QUOT(0', s(z0), s(z1)) -> c
   QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
   QUOT(z0, 0', s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))

The (relative) TRS S consists of the following rules:

   quot(0', s(z0), s(z1)) -> 0'
   quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
   quot(z0, 0', s(z1)) -> s(quot(z0, s(z1), s(z1)))

Rewrite Strategy: INNERMOST
----------------------------------------

(17) TypeInferenceProof (BOTH BOUNDS(ID, ID))
Inferred types.
----------------------------------------

(18)
Obligation:
Innermost TRS:
Rules:
QUOT(0', s(z0), s(z1)) -> c
QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
QUOT(z0, 0', s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
quot(0', s(z0), s(z1)) -> 0'
quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
quot(z0, 0', s(z1)) -> s(quot(z0, s(z1), s(z1)))

Types:
QUOT :: 0':s -> 0':s -> 0':s -> c:c1:c2
0' :: 0':s
s :: 0':s -> 0':s
c :: c:c1:c2
c1 :: c:c1:c2 -> c:c1:c2
c2 :: c:c1:c2 -> c:c1:c2
quot :: 0':s -> 0':s -> 0':s -> 0':s
hole_c:c1:c21_3 :: c:c1:c2
hole_0':s2_3 :: 0':s
gen_c:c1:c23_3 :: Nat -> c:c1:c2
gen_0':s4_3 :: Nat -> 0':s

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

(19) OrderProof (LOWER BOUND(ID))
Heuristically decided to analyse the following defined symbols:
QUOT, quot
----------------------------------------

(20)
Obligation:
Innermost TRS:
Rules:
QUOT(0', s(z0), s(z1)) -> c
QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
QUOT(z0, 0', s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
quot(0', s(z0), s(z1)) -> 0'
quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
quot(z0, 0', s(z1)) -> s(quot(z0, s(z1), s(z1)))

Types:
QUOT :: 0':s -> 0':s -> 0':s -> c:c1:c2
0' :: 0':s
s :: 0':s -> 0':s
c :: c:c1:c2
c1 :: c:c1:c2 -> c:c1:c2
c2 :: c:c1:c2 -> c:c1:c2
quot :: 0':s -> 0':s -> 0':s -> 0':s
hole_c:c1:c21_3 :: c:c1:c2
hole_0':s2_3 :: 0':s
gen_c:c1:c23_3 :: Nat -> c:c1:c2
gen_0':s4_3 :: Nat -> 0':s


Generator Equations:
gen_c:c1:c23_3(0) <=> c
gen_c:c1:c23_3(+(x, 1)) <=> c1(gen_c:c1:c23_3(x))
gen_0':s4_3(0) <=> 0'
gen_0':s4_3(+(x, 1)) <=> s(gen_0':s4_3(x))


The following defined symbols remain to be analysed:
QUOT, quot
----------------------------------------

(21) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
QUOT(gen_0':s4_3(n6_3), gen_0':s4_3(+(1, n6_3)), gen_0':s4_3(1)) -> gen_c:c1:c23_3(n6_3), rt in Omega(1 + n6_3)

Induction Base:
QUOT(gen_0':s4_3(0), gen_0':s4_3(+(1, 0)), gen_0':s4_3(1)) ->_R^Omega(1)
c

Induction Step:
QUOT(gen_0':s4_3(+(n6_3, 1)), gen_0':s4_3(+(1, +(n6_3, 1))), gen_0':s4_3(1)) ->_R^Omega(1)
c1(QUOT(gen_0':s4_3(n6_3), gen_0':s4_3(+(1, n6_3)), gen_0':s4_3(1))) ->_IH
c1(gen_c:c1:c23_3(c7_3))

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

(22)
Complex Obligation (BEST)

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

(23)
Obligation:
Proved the lower bound n^1 for the following obligation:

Innermost TRS:
Rules:
QUOT(0', s(z0), s(z1)) -> c
QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
QUOT(z0, 0', s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
quot(0', s(z0), s(z1)) -> 0'
quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
quot(z0, 0', s(z1)) -> s(quot(z0, s(z1), s(z1)))

Types:
QUOT :: 0':s -> 0':s -> 0':s -> c:c1:c2
0' :: 0':s
s :: 0':s -> 0':s
c :: c:c1:c2
c1 :: c:c1:c2 -> c:c1:c2
c2 :: c:c1:c2 -> c:c1:c2
quot :: 0':s -> 0':s -> 0':s -> 0':s
hole_c:c1:c21_3 :: c:c1:c2
hole_0':s2_3 :: 0':s
gen_c:c1:c23_3 :: Nat -> c:c1:c2
gen_0':s4_3 :: Nat -> 0':s


Generator Equations:
gen_c:c1:c23_3(0) <=> c
gen_c:c1:c23_3(+(x, 1)) <=> c1(gen_c:c1:c23_3(x))
gen_0':s4_3(0) <=> 0'
gen_0':s4_3(+(x, 1)) <=> s(gen_0':s4_3(x))


The following defined symbols remain to be analysed:
QUOT, quot
----------------------------------------

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

(25)
BOUNDS(n^1, INF)

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

(26)
Obligation:
Innermost TRS:
Rules:
QUOT(0', s(z0), s(z1)) -> c
QUOT(s(z0), s(z1), z2) -> c1(QUOT(z0, z1, z2))
QUOT(z0, 0', s(z1)) -> c2(QUOT(z0, s(z1), s(z1)))
quot(0', s(z0), s(z1)) -> 0'
quot(s(z0), s(z1), z2) -> quot(z0, z1, z2)
quot(z0, 0', s(z1)) -> s(quot(z0, s(z1), s(z1)))

Types:
QUOT :: 0':s -> 0':s -> 0':s -> c:c1:c2
0' :: 0':s
s :: 0':s -> 0':s
c :: c:c1:c2
c1 :: c:c1:c2 -> c:c1:c2
c2 :: c:c1:c2 -> c:c1:c2
quot :: 0':s -> 0':s -> 0':s -> 0':s
hole_c:c1:c21_3 :: c:c1:c2
hole_0':s2_3 :: 0':s
gen_c:c1:c23_3 :: Nat -> c:c1:c2
gen_0':s4_3 :: Nat -> 0':s


Lemmas:
QUOT(gen_0':s4_3(n6_3), gen_0':s4_3(+(1, n6_3)), gen_0':s4_3(1)) -> gen_c:c1:c23_3(n6_3), rt in Omega(1 + n6_3)


Generator Equations:
gen_c:c1:c23_3(0) <=> c
gen_c:c1:c23_3(+(x, 1)) <=> c1(gen_c:c1:c23_3(x))
gen_0':s4_3(0) <=> 0'
gen_0':s4_3(+(x, 1)) <=> s(gen_0':s4_3(x))


The following defined symbols remain to be analysed:
quot
----------------------------------------

(27) RewriteLemmaProof (LOWER BOUND(ID))
Proved the following rewrite lemma:
quot(gen_0':s4_3(n2224_3), gen_0':s4_3(+(1, n2224_3)), gen_0':s4_3(1)) -> gen_0':s4_3(0), rt in Omega(0)

Induction Base:
quot(gen_0':s4_3(0), gen_0':s4_3(+(1, 0)), gen_0':s4_3(1)) ->_R^Omega(0)
0'

Induction Step:
quot(gen_0':s4_3(+(n2224_3, 1)), gen_0':s4_3(+(1, +(n2224_3, 1))), gen_0':s4_3(1)) ->_R^Omega(0)
quot(gen_0':s4_3(n2224_3), gen_0':s4_3(+(1, n2224_3)), gen_0':s4_3(1)) ->_IH
gen_0':s4_3(0)

We have rt in Omega(1) and sz in O(n). Thus, we have irc_R in Omega(n^0).
----------------------------------------

(28)
BOUNDS(1, INF)