Termination Proof using AProVE

Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IF_GCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
IF_GCD(true, s(x), s(y)) -> MINUS(x, y)
IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IF_GCD(false, s(x), s(y)) -> MINUS(y, x)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS 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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

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

         ↳Nar

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS 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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Narrowing Transformation

Dependency Pairs:

IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IF_GCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

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)) -> IF_GCD(le(y, x), s(x), s(y))

three new Dependency Pairs are created:

GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))
GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))
IF_GCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))
IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

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

IF_GCD(true, s(x), s(y)) -> GCD(minus(x, y), s(y))

two new Dependency Pairs are created:

IF_GCD(true, s(x''), s(0)) -> GCD(x'', s(0))
IF_GCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pairs:

IF_GCD(true, s(x''), s(0)) -> GCD(x'', s(0))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF_GCD(true, s(x''), s(0)) -> GCD(x'', s(0))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))

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

{GCD, 0, IF_GCD} > s

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

GCD(x₁, x₂) -> GCD(x₁, x₂)
IF_GCD(x₁, x₂, x₃) -> IF_GCD(x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Dependency Graph

Dependency Pair:

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

IF_GCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))
GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))
IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

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

IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))

two new Dependency Pairs are created:

IF_GCD(false, s(0), s(y')) -> GCD(y', s(0))
IF_GCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Instantiation Transformation

Dependency Pairs:

IF_GCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_GCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

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

IF_GCD(true, s(x''), s(s(y''))) -> GCD(pred(minus(x'', y'')), s(s(y'')))

one new Dependency Pair is created:

IF_GCD(true, s(s(y'''')), s(s(y''0))) -> GCD(pred(minus(s(y''''), y''0)), s(s(y''0)))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Instantiation Transformation

Dependency Pairs:

IF_GCD(true, s(s(y'''')), s(s(y''0))) -> GCD(pred(minus(s(y''''), y''0)), s(s(y''0)))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_GCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

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

IF_GCD(false, s(s(y'')), s(y0)) -> GCD(pred(minus(y0, y'')), s(s(y'')))

one new Dependency Pair is created:

IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(pred(minus(s(x''''), y'''')), s(s(y'''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(pred(minus(s(x''''), y'''')), s(s(y'''')))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_GCD(true, s(s(y'''')), s(s(y''0))) -> GCD(pred(minus(s(y''''), y''0)), s(s(y''0)))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(x))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:46 minutes