Termination Proof using AProVE

Term Rewriting System R:
[x, y]
f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(x)) -> F(x)
G(s(x), s(y)) -> IF(f(x), s(x), s(y))
G(s(x), s(y)) -> F(x)
G(x, c(y)) -> G(x, g(s(c(y)), y))
G(x, c(y)) -> G(s(c(y)), y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

F(s(x)) -> F(x)

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

F(s(x)) -> F(x)

one new Dependency Pair is created:

F(s(s(x''))) -> F(s(x''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Nar

Dependency Pair:

F(s(s(x''))) -> F(s(x''))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

F(s(s(x''))) -> F(s(x''))

one new Dependency Pair is created:

F(s(s(s(x'''')))) -> F(s(s(x'''')))

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

         ↳Nar

Dependency Pair:

F(s(s(s(x'''')))) -> F(s(s(x'''')))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(s(s(s(x'''')))) -> F(s(s(x'''')))

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(F(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

G(x, c(y)) -> G(s(c(y)), y)
G(x, c(y)) -> G(x, g(s(c(y)), y))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

G(x, c(y)) -> G(x, g(s(c(y)), y))

two new Dependency Pairs are created:

G(x, c(s(y''))) -> G(x, if(f(c(s(y''))), s(c(s(y''))), s(y'')))
G(x, c(c(y''))) -> G(x, g(s(c(c(y''))), g(s(c(y'')), y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

G(x, c(c(y''))) -> G(x, g(s(c(c(y''))), g(s(c(y'')), y'')))
G(x, c(s(y''))) -> G(x, if(f(c(s(y''))), s(c(s(y''))), s(y'')))
G(x, c(y)) -> G(s(c(y)), y)

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

G(x, c(s(y''))) -> G(x, if(f(c(s(y''))), s(c(s(y''))), s(y'')))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

G(x, c(y)) -> G(s(c(y)), y)
G(x, c(c(y''))) -> G(x, g(s(c(c(y''))), g(s(c(y'')), y'')))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

G(x, c(c(y''))) -> G(x, g(s(c(c(y''))), g(s(c(y'')), y'')))

two new Dependency Pairs are created:

G(x, c(c(s(y')))) -> G(x, g(s(c(c(s(y')))), if(f(c(s(y'))), s(c(s(y'))), s(y'))))
G(x, c(c(c(y')))) -> G(x, g(s(c(c(c(y')))), g(s(c(c(y'))), g(s(c(y')), y'))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

G(x, c(c(c(y')))) -> G(x, g(s(c(c(c(y')))), g(s(c(c(y'))), g(s(c(y')), y'))))
G(x, c(c(s(y')))) -> G(x, g(s(c(c(s(y')))), if(f(c(s(y'))), s(c(s(y'))), s(y'))))
G(x, c(y)) -> G(s(c(y)), y)

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

G(x, c(c(s(y')))) -> G(x, g(s(c(c(s(y')))), if(f(c(s(y'))), s(c(s(y'))), s(y'))))

no new Dependency Pairs are created.
The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(x, c(y)) -> G(s(c(y)), y)
G(x, c(c(c(y')))) -> G(x, g(s(c(c(c(y')))), g(s(c(c(y'))), g(s(c(y')), y'))))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

G(x, c(y)) -> G(s(c(y)), y)

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Forward Instantiation Transformation

Dependency Pairs:

G(x, c(c(c(c(y'''))))) -> G(s(c(c(c(c(y'''))))), c(c(c(y'''))))
G(x, c(c(y''))) -> G(s(c(c(y''))), c(y''))
G(x, c(c(c(y')))) -> G(x, g(s(c(c(c(y')))), g(s(c(c(y'))), g(s(c(y')), y'))))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Polynomial Ordering

Dependency Pairs:

G(x, c(c(c(c(c(y''''')))))) -> G(s(c(c(c(c(c(y''''')))))), c(c(c(c(y''''')))))
G(x, c(c(c(y'''')))) -> G(s(c(c(c(y'''')))), c(c(y'''')))
G(x, c(c(c(c(y''''))))) -> G(s(c(c(c(c(y''''))))), c(c(c(y''''))))
G(x, c(c(c(y')))) -> G(x, g(s(c(c(c(y')))), g(s(c(c(y'))), g(s(c(y')), y'))))
G(x, c(c(c(c(y'''))))) -> G(s(c(c(c(c(y'''))))), c(c(c(y'''))))

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

G(x, c(c(c(y')))) -> G(x, g(s(c(c(c(y')))), g(s(c(c(y'))), g(s(c(y')), y'))))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))
if(true, x, y) -> x
if(false, x, y) -> y

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(if(x₁, x₂, x₃)) = x₂ + x₃

POL(c(x₁)) = 1

POL(0) = 0

POL(g(x₁, x₂)) = 0

POL(false) = 0

POL(G(x₁, x₂)) = x₂

POL(1) = 0

POL(true) = 0

POL(s(x₁)) = 0

POL(f(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Instantiation Transformation

Dependency Pairs:

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

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

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

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

two new Dependency Pairs are created:

G(s(c(c(c(c(c(y''''')))))), c(c(c(c(y'''''))))) -> G(s(c(c(c(c(y'''''))))), c(c(c(y'''''))))
G(s(c(c(c(c(c(y'''')))))), c(c(c(c(y''''))))) -> G(s(c(c(c(c(y''''))))), c(c(c(y''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 13

                 ↳Polynomial Ordering

Dependency Pairs:

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

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(c(x₁)) = 1 + x₁

POL(G(x₁, x₂)) = 1 + x₂

POL(s(x₁)) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 14

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(0) -> true
f(1) -> false
f(s(x)) -> f(x)
if(true, x, y) -> x
if(false, x, y) -> y
g(s(x), s(y)) -> if(f(x), s(x), s(y))
g(x, c(y)) -> g(x, g(s(c(y)), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(if(x₁, x₂, x₃))	= x₂ + x₃
POL(c(x₁))	= 1
POL(0)	= 0
POL(g(x₁, x₂))	= 0
POL(false)	= 0
POL(G(x₁, x₂))	= x₂
POL(1)	= 0
POL(true)	= 0
POL(s(x₁))	= 0
POL(f(x₁))	= 0

POL(c(x₁))	= 1 + x₁
POL(G(x₁, x₂))	= 1 + x₂
POL(s(x₁))	= 0