Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))
MINUS(s(x), s(y)) -> P(s(x))
MINUS(s(x), s(y)) -> P(s(y))
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(x, plus(y, z)) -> MINUS(x, y)
P(s(s(x))) -> P(s(x))
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))
DIV(s(x), s(y)) -> MINUS(x, y)
DIV(plus(x, y), z) -> PLUS(div(x, z), div(y, z))
DIV(plus(x, y), z) -> DIV(x, z)
DIV(plus(x, y), z) -> DIV(y, z)
PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))
PLUS(s(x), y) -> MINUS(s(x), s(0))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pair:

P(s(s(x))) -> P(s(x))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

P(s(s(x))) -> P(s(x))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(P(x₁)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

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

P(x₁) -> P(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

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

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)
MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

MINUS(x, plus(y, z)) -> MINUS(x, y)
MINUS(x, plus(y, z)) -> MINUS(minus(x, y), z)

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

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(plus(x₁, x₂)) = 1 + x₁ + x₂

POL(0) = 0

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(minus(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = x₁

POL(p(x₁)) = x₁

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

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
plus(x₁, x₂) -> plus(x₁, x₂)
minus(x₁, x₂) -> minus(x₁, x₂)
s(x₁) -> s(x₁)
p(x₁) -> p(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Narrowing Transformation

       →DP Problem 3

         ↳Remaining

       →DP Problem 4

         ↳Remaining

Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:

innermost

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

MINUS(s(x), s(y)) -> MINUS(p(s(x)), p(s(y)))

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MINUS(s(x), s(s(x''))) -> MINUS(p(s(x)), s(p(s(x''))))
MINUS(s(s(x'')), s(y)) -> MINUS(s(p(s(x''))), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost
Dependency Pair:
PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost
Dependency Pairs:
DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MINUS(s(x), s(s(x''))) -> MINUS(p(s(x)), s(p(s(x''))))
MINUS(s(s(x'')), s(y)) -> MINUS(s(p(s(x''))), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost
Dependency Pair:
PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost
Dependency Pairs:
DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Remaining Obligation(s)

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MINUS(s(x), s(s(x''))) -> MINUS(p(s(x)), s(p(s(x''))))
MINUS(s(s(x'')), s(y)) -> MINUS(s(p(s(x''))), p(s(y)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost
Dependency Pair:
PLUS(s(x), y) -> PLUS(y, minus(s(x), s(0)))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost
Dependency Pairs:
DIV(plus(x, y), z) -> DIV(y, z)
DIV(plus(x, y), z) -> DIV(x, z)
DIV(s(x), s(y)) -> DIV(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(0, y) -> 0
minus(s(x), s(y)) -> minus(p(s(x)), p(s(y)))
minus(x, plus(y, z)) -> minus(minus(x, y), z)
p(s(s(x))) -> s(p(s(x)))
p(0) -> s(s(0))
div(s(x), s(y)) -> s(div(minus(x, y), s(y)))
div(plus(x, y), z) -> plus(div(x, z), div(y, z))
plus(0, y) -> y
plus(s(x), y) -> s(plus(y, minus(s(x), s(0))))

Strategy:
innermost

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

POL(plus(x₁, x₂))	= 1 + x₁ + x₂
POL(0)	= 0
POL(MINUS(x₁, x₂))	= x₁ + x₂
POL(minus(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= x₁
POL(p(x₁))	= x₁