(0) Obligation:

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

minus(X, 0) → X
minus(s(X), s(Y)) → p(minus(X, Y))
p(s(X)) → X
div(0, s(Y)) → 0
div(s(X), s(Y)) → s(div(minus(X, Y), s(Y)))

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(P(minus(z0, z1)), MINUS(z0, z1))
DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

MINUS, DIV

Compound Symbols:

c1, c4

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:none
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

(5) CdtPolyRedPairProof (UPPER BOUND (ADD(O(n^1))) transformation)

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
We considered the (Usable) Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(MINUS(x1, x2)) = 0   
POL(c1(x1)) = x1   
POL(c4(x1, x2)) = x1 + x2   
POL(minus(x1, x2)) = x1   
POL(p(x1)) = [1] + x1   
POL(s(x1)) = [1] + x1   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
K tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

(7) CdtPolyRedPairProof (UPPER BOUND (ADD(O(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))
We considered the (Usable) Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = [2]x12   
POL(MINUS(x1, x2)) = [1] + [2]x1   
POL(c1(x1)) = x1   
POL(c4(x1, x2)) = x1 + x2   
POL(minus(x1, x2)) = x1   
POL(p(x1)) = [1] + x1   
POL(s(x1)) = [1] + x1   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
S tuples:none
K tuples:

DIV(s(z0), s(z1)) → c4(DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

DIV, MINUS

Compound Symbols:

c4, c1

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(10) BOUNDS(O(1), O(1))