Termination Proof using AProVE

Term Rewriting System R:
[x, y]
f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(s(x))) -> P(h(g(x)))
F(s(s(x))) -> H(g(x))
F(s(s(x))) -> G(x)
F(s(s(x))) -> +'(p(g(x)), q(g(x)))
F(s(s(x))) -> P(g(x))
F(s(s(x))) -> Q(g(x))
G(s(x)) -> H(g(x))
G(s(x)) -> G(x)
G(s(x)) -> +'(p(g(x)), q(g(x)))
G(s(x)) -> P(g(x))
G(s(x)) -> Q(g(x))
H(x) -> +'(p(x), q(x))
H(x) -> P(x)
H(x) -> Q(x)
+'(x, s(y)) -> +'(x, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

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

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

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

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

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

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(+'(x₁, x₂)) = 1 + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

Dependency Pair:

G(s(x)) -> G(x)

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

G(s(x)) -> G(x)

one new Dependency Pair is created:

G(s(s(x''))) -> G(s(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Forward Instantiation Transformation

Dependency Pair:

G(s(s(x''))) -> G(s(x''))

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

G(s(s(x''))) -> G(s(x''))

one new Dependency Pair is created:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pair:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(s(s(s(x'''')))) -> G(s(s(x'''')))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(G(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(0) -> 0
f(s(0)) -> s(0)
f(s(s(x))) -> p(h(g(x)))
f(s(s(x))) -> +(p(g(x)), q(g(x)))
g(0) -> pair(s(0), s(0))
g(s(x)) -> h(g(x))
g(s(x)) -> pair(+(p(g(x)), q(g(x))), p(g(x)))
h(x) -> pair(+(p(x), q(x)), p(x))
p(pair(x, y)) -> x
q(pair(x, y)) -> y
+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes