Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
*'(x, s(y)) -> *'(x, y)
ODD(s(s(x))) -> ODD(x)
HALF(s(s(x))) -> HALF(x)
POW(x, y) -> F(x, y, s(0))
F(x, s(y), z) -> IF(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))
F(x, s(y), z) -> ODD(s(y))
F(x, s(y), z) -> F(x, y, *(x, z))
F(x, s(y), z) -> *'(x, z)
F(x, s(y), z) -> F(*(x, x), half(s(y)), z)
F(x, s(y), z) -> *'(x, x)
F(x, s(y), z) -> HALF(s(y))

Furthermore, R contains five SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

-'(s(x), s(y)) -> -'(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

-'(s(x), s(y)) -> -'(x, y)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(-'(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

-'(x₁, x₂) -> -'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

*'(x, s(y)) -> *'(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

*'(x, s(y)) -> *'(x, y)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(*'(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

*'(x₁, x₂) -> *'(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

ODD(s(s(x))) -> ODD(x)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

ODD(s(s(x))) -> ODD(x)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(ODD(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

ODD(x₁) -> ODD(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 8

             ↳Dependency Graph

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Argument Filtering and Ordering

       →DP Problem 5

         ↳AFS

Dependency Pair:

HALF(s(s(x))) -> HALF(x)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

HALF(s(s(x))) -> HALF(x)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(HALF(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

HALF(x₁) -> HALF(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

           →DP Problem 9

             ↳Dependency Graph

       →DP Problem 5

         ↳AFS

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳Argument Filtering and Ordering

Dependency Pairs:

F(x, s(y), z) -> F(*(x, x), half(s(y)), z)
F(x, s(y), z) -> F(x, y, *(x, z))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(x, s(y), z) -> F(x, y, *(x, z))

The following usable rules for innermost can be oriented:

*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(s(x₁)) = 1 + x₁

POL(half(x₁)) = x₁

POL(F(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)
*(x₁, x₂) -> x₂
half(x₁) -> half(x₁)
+(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳AFS

       →DP Problem 5

         ↳AFS

           →DP Problem 10

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

F(x, s(y), z) -> F(*(x, x), half(s(y)), z)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
*(x, 0) -> 0
*(x, s(y)) -> +(*(x, y), x)
if(true, x, y) -> x
if(false, x, y) -> y
if(true, x, y) -> true
if(false, x, y) -> false
odd(0) -> false
odd(s(0)) -> true
odd(s(s(x))) -> odd(x)
half(0) -> 0
half(s(0)) -> 0
half(s(s(x))) -> s(half(x))
pow(x, y) -> f(x, y, s(0))
f(x, 0, z) -> z
f(x, s(y), z) -> if(odd(s(y)), f(x, y, *(x, z)), f(*(x, x), half(s(y)), z))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:03 minutes

POL(-'(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(0)	= 0
POL(s(x₁))	= 1 + x₁
POL(half(x₁))	= x₁
POL(F(x₁, x₂, x₃))	= 1 + x₁ + x₂ + x₃