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

0 CpxTRS
↳1 NestedDefinedSymbolProof (BOTH BOUNDS(ID, ID), 0 ms)
↳2 CpxTRS
↳3 CpxTrsToCdtProof (BOTH BOUNDS(ID, ID), 0 ms)
↳4 CdtProblem
↳5 CdtLeafRemovalProof (BOTH BOUNDS(ID, ID), 0 ms)
↳6 CdtProblem
↳7 CdtUsableRulesProof (⇔, 0 ms)
↳8 CdtProblem
↳9 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 138 ms)
↳10 CdtProblem
↳11 CdtKnowledgeProof (BOTH BOUNDS(ID, ID), 0 ms)
↳12 CdtProblem
↳13 CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)), 16 ms)
↳14 CdtProblem
↳15 CdtRuleRemovalProof (UPPER BOUND(ADD(n^2)), 94 ms)
↳16 CdtProblem
↳17 SIsEmptyProof (BOTH BOUNDS(ID, ID), 0 ms)
↳18 BOUNDS(1, 1)

(0) Obligation:

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


The TRS R consists of the following rules:

plus(x, 0) → x
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
times(0, y) → 0
times(s(0), y) → y
times(s(x), y) → plus(y, times(x, y))
div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))
div(div(x, y), z) → div(x, times(y, z))

Rewrite Strategy: INNERMOST

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

The following defined symbols can occur below the 1th argument of plus: plus, times

Hence, the left-hand sides of the following rules are not basic-reachable and can be removed:
div(div(x, y), z) → div(x, times(y, z))

(2) Obligation:

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


The TRS R consists of the following rules:

plus(s(x), y) → s(plus(x, y))
times(s(x), y) → plus(y, times(x, y))
quot(x, 0, s(z)) → s(div(x, s(z)))
plus(x, 0) → x
plus(0, y) → y
div(0, y) → 0
div(x, y) → quot(x, y, y)
times(s(0), y) → y
quot(s(x), s(y), z) → quot(x, y, z)
times(0, y) → 0
quot(0, s(y), z) → 0

Rewrite Strategy: INNERMOST

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

Converted Cpx (relative) TRS to CDT

(4) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(0, s(z0), z1) → 0
div(0, z0) → 0
div(z0, z1) → quot(z0, z1, z1)
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
PLUS(z0, 0) → c1
PLUS(0, z0) → c2
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
TIMES(s(0), z0) → c4
TIMES(0, z0) → c5
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(0, s(z0), z1) → c8
DIV(0, z0) → c9
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
PLUS(z0, 0) → c1
PLUS(0, z0) → c2
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
TIMES(s(0), z0) → c4
TIMES(0, z0) → c5
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(0, s(z0), z1) → c8
DIV(0, z0) → c9
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:none
Defined Rule Symbols:

plus, times, quot, div

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c1, c2, c3, c4, c5, c6, c7, c8, c9, c10

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

Removed 6 trailing nodes:

PLUS(0, z0) → c2
QUOT(0, s(z0), z1) → c8
TIMES(s(0), z0) → c4
TIMES(0, z0) → c5
DIV(0, z0) → c9
PLUS(z0, 0) → c1

(6) Obligation:

Complexity Dependency Tuples Problem
Rules:

plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(0, s(z0), z1) → 0
div(0, z0) → 0
div(z0, z1) → quot(z0, z1, z1)
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:none
Defined Rule Symbols:

plus, times, quot, div

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

(7) CdtUsableRulesProof (EQUIVALENT transformation)

The following rules are not usable and were removed:

quot(z0, 0, s(z1)) → s(div(z0, s(z1)))
quot(s(z0), s(z1), z2) → quot(z0, z1, z2)
quot(0, s(z0), z1) → 0
div(0, z0) → 0
div(z0, z1) → quot(z0, z1, z1)

(8) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:none
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

(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.

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
We considered the (Usable) Rules:none
And the Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = x1   
POL(PLUS(x1, x2)) = 0   
POL(QUOT(x1, x2, x3)) = x1   
POL(TIMES(x1, x2)) = 0   
POL(c(x1)) = x1   
POL(c10(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = [1] + x1   
POL(times(x1, x2)) = 0   

(10) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

(11) CdtKnowledgeProof (BOTH BOUNDS(ID, ID) transformation)

The following tuples could be moved from S to K by knowledge propagation:

QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))

(12) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

(13) CdtRuleRemovalProof (UPPER BOUND(ADD(n^1)) transformation)

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

TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = 0   
POL(DIV(x1, x2)) = 0   
POL(PLUS(x1, x2)) = 0   
POL(QUOT(x1, x2, x3)) = 0   
POL(TIMES(x1, x2)) = x1   
POL(c(x1)) = x1   
POL(c10(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = [1] + x1   
POL(times(x1, x2)) = 0   

(14) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

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

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

PLUS(s(z0), z1) → c(PLUS(z0, z1))
We considered the (Usable) Rules:none
And the Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
The order we found is given by the following interpretation:
Polynomial interpretation :

POL(0) = [1]   
POL(DIV(x1, x2)) = 0   
POL(PLUS(x1, x2)) = x1   
POL(QUOT(x1, x2, x3)) = 0   
POL(TIMES(x1, x2)) = [2]x1·x2   
POL(c(x1)) = x1   
POL(c10(x1)) = x1   
POL(c3(x1, x2)) = x1 + x2   
POL(c6(x1)) = x1   
POL(c7(x1)) = x1   
POL(plus(x1, x2)) = [2] + [2]x1 + [2]x2 + x22 + [2]x1·x2 + [2]x12   
POL(s(x1)) = [2] + x1   
POL(times(x1, x2)) = [1] + x1 + x2 + x22 + x1·x2 + x12   

(16) Obligation:

Complexity Dependency Tuples Problem
Rules:

times(s(z0), z1) → plus(z1, times(z0, z1))
times(s(0), z0) → z0
times(0, z0) → 0
plus(s(z0), z1) → s(plus(z0, z1))
plus(z0, 0) → z0
plus(0, z0) → z0
Tuples:

PLUS(s(z0), z1) → c(PLUS(z0, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
S tuples:none
K tuples:

QUOT(s(z0), s(z1), z2) → c7(QUOT(z0, z1, z2))
QUOT(z0, 0, s(z1)) → c6(DIV(z0, s(z1)))
DIV(z0, z1) → c10(QUOT(z0, z1, z1))
TIMES(s(z0), z1) → c3(PLUS(z1, times(z0, z1)), TIMES(z0, z1))
PLUS(s(z0), z1) → c(PLUS(z0, z1))
Defined Rule Symbols:

times, plus

Defined Pair Symbols:

PLUS, TIMES, QUOT, DIV

Compound Symbols:

c, c3, c6, c7, c10

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

The set S is empty

(18) BOUNDS(1, 1)