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

0 CpxTRS
↳1 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CdtProblem
↳3 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtUsableRulesProof (⇔, 0 ms)
↳6 CdtProblem
↳7 CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))), 52 ms)
↳8 CdtProblem
↳9 CdtKnowledgeProof (⇔, 0 ms)
↳10 BOUNDS(O(1), O(1))

(0) Obligation:

Runtime Complexity TRS:
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)))

Rewrite Strategy: INNERMOST

(1) CpxTrsToCdtProof (BOTH BOUNDS(ID, ID) transformation)

Converted Cpx (relative) TRS 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

Defined Pair Symbols:

QUOT

Compound Symbols:

c, c1, c2

(3) CdtLeafRemovalProof (BOTH BOUNDS(ID, ID) transformation)

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

Defined Pair Symbols:

QUOT

Compound Symbols:

c1, c2

(5) CdtUsableRulesProof (EQUIVALENT transformation)

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

Compound Symbols:

c1, c2

(7) CdtRuleRemovalProof (UPPER BOUND (ADD(O(n^1))) transformation)

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(x1, x2, x3)) = [5]x1   
POL(c1(x1)) = x1   
POL(c2(x1)) = x1   
POL(s(x1)) = [1] + x1   

(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

Compound Symbols:

c1, c2

(9) CdtKnowledgeProof (EQUIVALENT transformation)

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(O(1), O(1))