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

(0) Obligation:

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

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

Rewrite Strategy: INNERMOST

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

Converted CpxTRS to CDT

(2) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
geq(z0, 0) → true
geq(0, s(z0)) → false
geq(s(z0), s(z1)) → geq(z0, z1)
div(0, s(z0)) → 0
div(s(z0), s(z1)) → if(geq(z0, z1), s(div(minus(z0, z1), s(z1))), 0)
if(true, z0, z1) → z0
if(false, z0, z1) → z1

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(IF(geq(z0, z1), s(div(minus(z0, z1), s(z1))), 0), GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(IF(geq(z0, z1), s(div(minus(z0, z1), s(z1))), 0), GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

minus, geq, div, if

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

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

Removed 1 trailing tuple parts

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
geq(z0, 0) → true
geq(0, s(z0)) → false
geq(s(z0), s(z1)) → geq(z0, z1)
div(0, s(z0)) → 0
div(s(z0), s(z1)) → if(geq(z0, z1), s(div(minus(z0, z1), s(z1))), 0)
if(true, z0, z1) → z0
if(false, z0, z1) → z1

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

K tuples:none
Defined Rule Symbols:

minus, geq, div, if

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

(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.

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), 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)) = [2]x₁   
POL(GEQ(x₁, x₂)) = 0   
POL(MINUS(x₁, x₂)) = [2]x₁   
POL(c1(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(minus(x₁, x₂)) = 0   
POL(s(x₁)) = [1] + x₁   

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
geq(z0, 0) → true
geq(0, s(z0)) → false
geq(s(z0), s(z1)) → geq(z0, z1)
div(0, s(z0)) → 0
div(s(z0), s(z1)) → if(geq(z0, z1), s(div(minus(z0, z1), s(z1))), 0)
if(true, z0, z1) → z0
if(false, z0, z1) → z1

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:

GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))

K tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

Defined Rule Symbols:

minus, geq, div, if

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

(7) 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.

GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))

We considered the (Usable) Rules:

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

And the Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

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

POL(0) = [1]   
POL(DIV(x₁, x₂)) = x₁   
POL(GEQ(x₁, x₂)) = x₁   
POL(MINUS(x₁, x₂)) = 0   
POL(c1(x₁)) = x₁   
POL(c4(x₁)) = x₁   
POL(c6(x₁, x₂, x₃)) = x₁ + x₂ + x₃   
POL(minus(x₁, x₂)) = [1]   
POL(s(x₁)) = [1] + x₁   

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

minus(0, z0) → 0
minus(s(z0), s(z1)) → minus(z0, z1)
geq(z0, 0) → true
geq(0, s(z0)) → false
geq(s(z0), s(z1)) → geq(z0, z1)
div(0, s(z0)) → 0
div(s(z0), s(z1)) → if(geq(z0, z1), s(div(minus(z0, z1), s(z1))), 0)
if(true, z0, z1) → z0
if(false, z0, z1) → z1

Tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))

S tuples:none
K tuples:

MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
DIV(s(z0), s(z1)) → c6(GEQ(z0, z1), DIV(minus(z0, z1), s(z1)), MINUS(z0, z1))
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))

Defined Rule Symbols:

minus, geq, div, if

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

(9) SIsEmptyProof (EQUIVALENT transformation)

The set S is empty

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