Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) -> 0
app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) -> x
app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0))))
APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))
APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(plus, y), app(app(times, app(s, z)), 0))
APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, app(s, z)), 0)
APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(times, app(s, z))
APP(app(times, x), app(app(plus, y), app(s, z))) -> APP(app(times, x), app(s, z))
APP(app(times, x), app(s, y)) -> APP(app(plus, app(app(times, x), y)), x)
APP(app(times, x), app(s, y)) -> APP(plus, app(app(times, x), y))
APP(app(times, x), app(s, y)) -> APP(app(times, x), y)
APP(app(plus, x), app(s, y)) -> APP(s, app(app(plus, x), y))
APP(app(plus, x), app(s, y)) -> APP(app(plus, x), y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

APP(app(plus, x), app(s, y)) -> APP(app(plus, x), y)

Rules:

app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) -> 0
app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) -> x
app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

APP(app(plus, x), app(s, y)) -> APP(app(plus, x), 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(plus) = 0

POL(0) = 0

POL(times) = 0

POL(s) = 0

POL(app(x₁, x₂)) = 1 + x₂

POL(APP(x₁, x₂)) = x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) -> 0
app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) -> x
app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

APP(app(times, x), app(s, y)) -> APP(app(times, x), y)
APP(app(times, x), app(s, y)) -> APP(app(plus, app(app(times, x), y)), x)

Rules:

app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) -> 0
app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) -> x
app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))

Strategy:

innermost

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

APP(app(times, x), app(s, y)) -> APP(app(plus, app(app(times, x), y)), x)

two new Dependency Pairs are created:

APP(app(times, x''), app(s, 0)) -> APP(app(plus, 0), x'')
APP(app(times, x''), app(s, app(s, y''))) -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), x'')), x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Narrowing Transformation

Dependency Pairs:

APP(app(times, x''), app(s, app(s, y''))) -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), x'')), x'')
APP(app(times, x''), app(s, 0)) -> APP(app(plus, 0), x'')
APP(app(times, x), app(s, y)) -> APP(app(times, x), y)

Rules:

app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) -> 0
app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) -> x
app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))

Strategy:

innermost

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

APP(app(times, x''), app(s, 0)) -> APP(app(plus, 0), x'')

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

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

APP(app(times, x), app(s, y)) -> APP(app(times, x), y)
APP(app(times, x''), app(s, app(s, y''))) -> APP(app(plus, app(app(plus, app(app(times, x''), y'')), x'')), x'')

Rules:

app(app(times, x), app(app(plus, y), app(s, z))) -> app(app(plus, app(app(times, x), app(app(plus, y), app(app(times, app(s, z)), 0)))), app(app(times, x), app(s, z)))
app(app(times, x), 0) -> 0
app(app(times, x), app(s, y)) -> app(app(plus, app(app(times, x), y)), x)
app(app(plus, x), 0) -> x
app(app(plus, x), app(s, y)) -> app(s, app(app(plus, x), y))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes

POL(plus)	= 0
POL(0)	= 0
POL(times)	= 0
POL(s)	= 0
POL(app(x₁, x₂))	= 1 + x₂
POL(APP(x₁, x₂))	= x₂