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

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

LE(s(s(x'')), s(s(y''))) -> LE(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 Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

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(LE(x₁, x₂)) = 1 + x₁ + x₂

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →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

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(x'', s(s(y''))) -> MINUS(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 Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

The following dependency pair can be strictly oriented:

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

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(MINUS(x₁, x₂)) = 1 + 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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 11

                 ↳Forward Instantiation Transformation

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 13

                 ↳Forward Instantiation Transformation

Dependency Pairs:

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

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

Dependency Pairs:

GCD(s(s(x'''''')), s(0)) -> IF_GCD(true, s(s(x'''''')), s(0))
IF_GCD(true, s(s(x'''')), s(0)) -> GCD(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 pair can be strictly oriented:

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

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(0) = 0

POL(GCD(x₁, x₂)) = x₁ + x₂

POL(IF_GCD(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(true) = 0

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 20

                 ↳Dependency Graph

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 12

                 ↳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

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 14

                 ↳Narrowing 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, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

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

three new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 16

                 ↳Narrowing Transformation

Dependency Pairs:

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

two new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 17

                 ↳Argument Filtering and Ordering

Dependency Pairs:

IF_GCD(true, s(x'''), s(s(0))) -> GCD(pred(x'''), s(s(0)))
GCD(s(s(y''')), s(s(0))) -> IF_GCD(true, s(s(y''')), s(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 pair can be strictly oriented:

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

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

pred(s(x)) -> x

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(GCD(x₁, x₂)) = x₁ + x₂

POL(pred(x₁)) = x₁

POL(IF_GCD(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(true) = 0

POL(s(x₁)) = 1 + x₁

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

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 21

                 ↳Dependency Graph

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 18

                 ↳Narrowing Transformation

Dependency Pairs:

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

two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 19

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

The following dependency pair can be strictly oriented:

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

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

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))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(GCD(x₁, x₂)) = 1 + x₁ + x₂

POL(false) = 0

POL(pred(x₁)) = x₁

POL(IF_GCD(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le) = 0

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

GCD(x₁, x₂) -> GCD(x₁, x₂)
IF_GCD(x₁, x₂, x₃) -> IF_GCD(x₁, x₂, x₃)
s(x₁) -> s(x₁)
le(x₁, x₂) -> le
pred(x₁) -> pred(x₁)
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

           →DP Problem 10

             ↳Nar

...

               →DP Problem 22

                 ↳Dependency Graph

Dependency Pairs:

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

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

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

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

POL(0)	= 0
POL(GCD(x₁, x₂))	= x₁ + x₂
POL(IF_GCD(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(true)	= 0
POL(s(x₁))	= 1 + x₁

POL(GCD(x₁, x₂))	= 1 + x₁ + x₂
POL(false)	= 0
POL(pred(x₁))	= x₁
POL(IF_GCD(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(le)	= 0