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

(0) Obligation:

The Runtime Complexity (full) 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: FULL

(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 (full) 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: FULL

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

Converted rc-obligation to irc-obligation.

The duplicating contexts are:
times(s(x), [])
div(x, [])

The defined contexts are:
plus(x0, [])

As the TRS is an overlay system and the defined contexts and the duplicating contexts don't overlap, we have rc = irc.

(4) 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

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

Converted Cpx (relative) TRS to CDT

(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))
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

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

Removed 6 trailing nodes:

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

(8) 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

(9) 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)

(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)))
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

(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.

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   

(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))
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

(13) 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))

(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))
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

(15) 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   

(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:

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

(17) 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   

(18) 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

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

The set S is empty

(20) BOUNDS(1, 1)