Termination Proof using AProVE

Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
MIN(s(x), s(y)) -> MIN(x, y)
TWICE(s(x)) -> TWICE(x)
F(s(x), s(y)) -> F(-(y, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) -> -'(y, min(x, y))
F(s(x), s(y)) -> MIN(x, y)
F(s(x), s(y)) -> TWICE(min(x, y))
F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) -> -'(x, min(x, y))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

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

       →DP Problem 4

         ↳Remaining

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

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

       →DP Problem 4

         ↳Remaining

Dependency Pair:

TWICE(s(x)) -> TWICE(x)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

TWICE(s(x)) -> TWICE(x)

There are no usable rules for innermost that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

TWICE(x₁) -> TWICE(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 7

             ↳Dependency Graph

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

F(s(x), s(y)) -> F(-(x, min(x, y)), s(twice(min(x, y))))
F(s(x), s(y)) -> F(-(y, min(x, y)), s(twice(min(x, y))))

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
twice(0) -> 0
twice(s(x)) -> s(s(twice(x)))
f(s(x), s(y)) -> f(-(y, min(x, y)), s(twice(min(x, y))))
f(s(x), s(y)) -> f(-(x, min(x, y)), s(twice(min(x, y))))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:04 minutes