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

(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) 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:

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))

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(P(x₁)) = 0   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = [3] + x₁   
POL(p(x₁)) = [3] + x₁   
POL(s(x₁)) = [5] + x₁   

(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:

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))

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:

MINUS, DIV

Compound Symbols:

c1, c4

(5) 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(P(minus(z0, 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:

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))

The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x₁, x₂)) = x₁·x₂ + [2]x₁²   
POL(MINUS(x₁, x₂)) = [3] + x₁ + [2]x₂   
POL(P(x₁)) = [1]   
POL(c1(x₁, x₂)) = x₁ + x₂   
POL(c4(x₁, x₂)) = x₁ + x₂   
POL(minus(x₁, x₂)) = [1] + x₁   
POL(p(x₁)) = x₁   
POL(s(x₁)) = [3] + 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:

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:none
K tuples:

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

Defined Rule Symbols:

minus, p, div

Defined Pair Symbols:

MINUS, DIV

Compound Symbols:

c1, c4

(7) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

(8) BOUNDS(O(1), O(1))