The Runtime Complexity (innermost) of the given CpxTRS could be proven to be BOUNDS(1, 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 CdtRhsSimplificationProcessorProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 125 ms)
↳10 CdtProblem
↳11 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 37 ms)
↳12 CdtProblem
↳13 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳14 BOUNDS(1, 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 Cpx (relative) TRS 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(0, z0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(z0, 0) → c2
GEQ(0, s(z0)) → c3
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(0, s(z0)) → c5
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))
IF(true, z0, z1) → c7
IF(false, z0, z1) → c8
S tuples:

MINUS(0, z0) → c
MINUS(s(z0), s(z1)) → c1(MINUS(z0, z1))
GEQ(z0, 0) → c2
GEQ(0, s(z0)) → c3
GEQ(s(z0), s(z1)) → c4(GEQ(z0, z1))
DIV(0, s(z0)) → c5
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))
IF(true, z0, z1) → c7
IF(false, z0, z1) → c8
K tuples:none
Defined Rule Symbols:

minus, geq, div, if

Defined Pair Symbols:

MINUS, GEQ, DIV, IF

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8

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

Removed 6 trailing nodes:

DIV(0, s(z0)) → c5
GEQ(0, s(z0)) → c3
MINUS(0, z0) → c
IF(false, z0, z1) → c8
IF(true, z0, z1) → c7
GEQ(z0, 0) → c2

(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(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

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

Removed 1 trailing tuple parts

(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:

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

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

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

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

(9) CdtRuleRemovalProof (UPPER BOUND(ADD(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(s(z0), s(z1)) → minus(z0, z1)
minus(0, z0) → 0
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   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

(11) CdtRuleRemovalProof (UPPER BOUND(ADD(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(s(z0), s(z1)) → minus(z0, z1)
minus(0, z0) → 0
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   

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

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

Defined Pair Symbols:

MINUS, GEQ, DIV

Compound Symbols:

c1, c4, c6

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

The set S is empty

(14) BOUNDS(1, 1)