Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MIN(s(x), s(y)) -> MIN(x, y)
MAX(s(x), s(y)) -> MAX(x, y)
-'(s(x), s(y)) -> -'(x, y)
GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))
GCD(s(x), s(y)) -> -'(s(max(x, y)), s(min(x, y)))
GCD(s(x), s(y)) -> MAX(x, y)
GCD(s(x), s(y)) -> 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

         ↳Nar

Dependency Pair:

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

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(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 w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
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 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Nar

Dependency Pair:

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(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

         ↳Nar

Dependency Pair:

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

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

MAX(x₁, x₂) -> MAX(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

         ↳Nar

Dependency Pair:

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(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

         ↳Nar

Dependency Pair:

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

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

-'(x₁, x₂) -> -'(x₁, x₂)
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

         ↳Nar

Dependency Pair:

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(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

         ↳Narrowing Transformation

Dependency Pair:

GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Strategy:

innermost

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

GCD(s(x), s(y)) -> GCD(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

10 new Dependency Pairs are created:

GCD(s(x''), s(y'')) -> GCD(-(max(x'', y''), min(x'', y'')), s(min(x'', y'')))
GCD(s(x''), s(0)) -> GCD(-(s(x''), s(min(x'', 0))), s(min(x'', 0)))
GCD(s(0), s(y'')) -> GCD(-(s(y''), s(min(0, y''))), s(min(0, y'')))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(s(max(x'', y''))), s(min(s(x''), s(y'')))), s(min(s(x''), s(y''))))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(0)), s(min(x'', 0)))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(0)), s(min(0, y'')))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(s(min(x'', y'')))), s(min(s(x''), s(y''))))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(min(x'', 0))), s(0))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(min(0, y''))), s(0))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(min(s(x''), s(y'')))), s(s(min(x'', y''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

       →DP Problem 4

         ↳Nar

           →DP Problem 8

             ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(min(s(x''), s(y'')))), s(s(min(x'', y''))))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(min(0, y''))), s(0))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(min(x'', 0))), s(0))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(max(s(x''), s(y''))), s(s(min(x'', y'')))), s(min(s(x''), s(y''))))
GCD(s(0), s(y'')) -> GCD(-(s(max(0, y'')), s(0)), s(min(0, y'')))
GCD(s(x''), s(0)) -> GCD(-(s(max(x'', 0)), s(0)), s(min(x'', 0)))
GCD(s(s(x'')), s(s(y''))) -> GCD(-(s(s(max(x'', y''))), s(min(s(x''), s(y'')))), s(min(s(x''), s(y''))))
GCD(s(0), s(y'')) -> GCD(-(s(y''), s(min(0, y''))), s(min(0, y'')))
GCD(s(x''), s(0)) -> GCD(-(s(x''), s(min(x'', 0))), s(min(x'', 0)))
GCD(s(x''), s(y'')) -> GCD(-(max(x'', y''), min(x'', y'')), s(min(x'', y'')))

Rules:

min(x, 0) -> 0
min(0, y) -> 0
min(s(x), s(y)) -> s(min(x, y))
max(x, 0) -> x
max(0, y) -> y
max(s(x), s(y)) -> s(max(x, y))
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
gcd(s(x), s(y)) -> gcd(-(s(max(x, y)), s(min(x, y))), s(min(x, y)))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:07 minutes