Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
plus(minus(x, s(0)), minus(y, s(s(z)))) -> plus(minus(y, s(s(z))), minus(x, s(0)))
plus(plus(x, s(0)), plus(y, s(s(z)))) -> plus(plus(y, s(s(z))), plus(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(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
PLUS(s(x), y) -> PLUS(x, y)
PLUS(minus(x, s(0)), minus(y, s(s(z)))) -> PLUS(minus(y, s(s(z))), minus(x, s(0)))
PLUS(plus(x, s(0)), plus(y, s(s(z)))) -> PLUS(plus(y, s(s(z))), plus(x, s(0)))

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

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

         ↳AFS

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

The following usable rules for innermost can be oriented:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
s(x₁) -> s(x₁)
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

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

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

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

POL(PLUS(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(MINUS(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(QUOT(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(PLUS(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁