The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, n^2).

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(n^2)), 126 ms)
↳8 CdtProblem
↳9 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳10 BOUNDS(1, 1)

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(z0, 0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(0, s(z0)) → c2
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, quot

Defined Pair Symbols:

MINUS, QUOT

Compound Symbols:

c, c1, c2, c3

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

Removed 2 trailing nodes:

MINUS(z0, 0) → c
QUOT(0, s(z0)) → c2

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, quot

Defined Pair Symbols:

MINUS, QUOT

Compound Symbols:

c1, c3

(5) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

quot(0, s(z0)) → 0
quot(s(z0), s(z1)) → s(quot(minus(z0, z1), s(z1)))

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS, QUOT

Compound Symbols:

c1, c3

(7) CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)) transformation)

Found a reduction pair which oriented the following tuples strictly. Hence they can be removed from S.

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

minus(s(z0), s(z1)) → minus(z0, z1)
minus(z0, 0) → z0
And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(MINUS(x1, x2)) = x1   
POL(QUOT(x1, x2)) = [2]x12   
POL(c1(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = [2] + x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(z0, 0) → z0
minus(s(z0), s(z1)) → minus(z0, z1)
Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:none
K tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
QUOT(s(z0), s(z1)) → c3(QUOT(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus

Defined Pair Symbols:

MINUS, QUOT

Compound Symbols:

c1, c3

(9) SIsEmptyProof (BOTH BOUNDS(ID, ID) transformation)

The set S is empty

(10) BOUNDS(1, 1)