Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:

MINUS(X, s(Y)) -> PRED(minus(X, Y))
MINUS(X, s(Y)) -> MINUS(X, Y)
LE(s(X), s(Y)) -> LE(X, Y)
GCD(s(X), s(Y)) -> IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) -> LE(Y, X)
IF(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))
IF(true, s(X), s(Y)) -> MINUS(X, Y)
IF(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))
IF(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:

MINUS(X, s(Y)) -> MINUS(X, Y)

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(X'', s(s(Y''))) -> MINUS(X'', s(Y''))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Nar

Dependency Pair:

MINUS(X'''', s(s(s(Y'''')))) -> MINUS(X'''', s(s(Y'''')))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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 that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = 1 + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:

LE(s(X), s(Y)) -> LE(X, Y)

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Nar

Dependency Pair:

LE(s(s(X'')), s(s(Y''))) -> LE(s(X''), s(Y''))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Nar

Dependency Pair:

LE(s(s(s(X''''))), s(s(s(Y'''')))) -> LE(s(s(X'''')), s(s(Y'''')))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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 that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳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:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))
IF(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))
GCD(s(X), s(Y)) -> IF(le(Y, X), s(X), s(Y))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(le(Y, X), s(X), s(Y))

three new Dependency Pairs are created:

GCD(s(s(Y'')), s(s(X''))) -> IF(le(X'', Y''), s(s(Y'')), s(s(X'')))
GCD(s(0), s(s(X''))) -> IF(false, s(0), s(s(X'')))
GCD(s(X'), s(0)) -> IF(true, 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

             ↳Narrowing Transformation

Dependency Pairs:

GCD(s(X'), s(0)) -> IF(true, s(X'), s(0))
GCD(s(0), s(s(X''))) -> IF(false, s(0), s(s(X'')))
IF(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))
GCD(s(s(Y'')), s(s(X''))) -> IF(le(X'', Y''), s(s(Y'')), s(s(X'')))
IF(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(true, s(X), s(Y)) -> GCD(minus(X, Y), s(Y))

two new Dependency Pairs are created:

IF(true, s(X''), s(s(Y''))) -> GCD(pred(minus(X'', Y'')), s(s(Y'')))
IF(true, s(X''), s(0)) -> GCD(X'', s(0))

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(true, s(X''), s(0)) -> GCD(X'', s(0))
GCD(s(X'), s(0)) -> IF(true, s(X'), s(0))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(true, s(X''), s(0)) -> GCD(X'', s(0))

one new Dependency Pair is created:

IF(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(true, s(s(X'''')), s(0)) -> GCD(s(X''''), s(0))
GCD(s(X'), s(0)) -> IF(true, s(X'), s(0))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(true, s(X'), s(0))

one new Dependency Pair is created:

GCD(s(s(X'''''')), s(0)) -> IF(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

                 ↳Polynomial Ordering

Dependency Pairs:

GCD(s(s(X'''''')), s(0)) -> IF(true, s(s(X'''''')), s(0))
IF(true, s(s(X'''')), s(0)) -> GCD(s(X''''), s(0))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF(true, s(s(X'''')), s(0)) -> GCD(s(X''''), s(0))

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(GCD(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(IF(x₁, x₂, x₃)) = x₂

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 20

                 ↳Dependency Graph

Dependency Pair:

GCD(s(s(X'''''')), s(0)) -> IF(true, s(s(X'''''')), s(0))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(true, s(X''), s(s(Y''))) -> GCD(pred(minus(X'', Y'')), s(s(Y'')))
GCD(s(s(Y'')), s(s(X''))) -> IF(le(X'', Y''), s(s(Y'')), s(s(X'')))
IF(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))
GCD(s(0), s(s(X''))) -> IF(false, s(0), s(s(X'')))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(false, s(X), s(Y)) -> GCD(minus(Y, X), s(X))

two new Dependency Pairs are created:

IF(false, s(s(Y'')), s(Y0)) -> GCD(pred(minus(Y0, Y'')), s(s(Y'')))
IF(false, s(0), s(Y')) -> GCD(Y', 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 14

                 ↳Narrowing Transformation

Dependency Pairs:

IF(false, s(s(Y'')), s(Y0)) -> GCD(pred(minus(Y0, Y'')), s(s(Y'')))
GCD(s(s(Y'')), s(s(X''))) -> IF(le(X'', Y''), s(s(Y'')), s(s(X'')))
IF(true, s(X''), s(s(Y''))) -> GCD(pred(minus(X'', Y'')), s(s(Y'')))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(le(X'', Y''), s(s(Y'')), s(s(X'')))

three new Dependency Pairs are created:

GCD(s(s(s(Y'))), s(s(s(X')))) -> IF(le(X', Y'), s(s(s(Y'))), s(s(s(X'))))
GCD(s(s(0)), s(s(s(X')))) -> IF(false, s(s(0)), s(s(s(X'))))
GCD(s(s(Y''')), s(s(0))) -> IF(true, s(s(Y''')), s(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 16

                 ↳Narrowing Transformation

Dependency Pairs:

GCD(s(s(Y''')), s(s(0))) -> IF(true, s(s(Y''')), s(s(0)))
GCD(s(s(0)), s(s(s(X')))) -> IF(false, s(s(0)), s(s(s(X'))))
IF(true, s(X''), s(s(Y''))) -> GCD(pred(minus(X'', Y'')), s(s(Y'')))
GCD(s(s(s(Y'))), s(s(s(X')))) -> IF(le(X', Y'), s(s(s(Y'))), s(s(s(X'))))
IF(false, s(s(Y'')), s(Y0)) -> GCD(pred(minus(Y0, Y'')), s(s(Y'')))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(true, s(X''), s(s(Y''))) -> GCD(pred(minus(X'', Y'')), s(s(Y'')))

two new Dependency Pairs are created:

IF(true, s(X'''), s(s(s(Y')))) -> GCD(pred(pred(minus(X''', Y'))), s(s(s(Y'))))
IF(true, s(X'''), s(s(0))) -> GCD(pred(X'''), s(s(0)))

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

                 ↳Polynomial Ordering

Dependency Pairs:

IF(true, s(X'''), s(s(0))) -> GCD(pred(X'''), s(s(0)))
GCD(s(s(Y''')), s(s(0))) -> IF(true, s(s(Y''')), s(s(0)))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF(true, s(X'''), s(s(0))) -> GCD(pred(X'''), s(s(0)))

Additionally, the following usable rule for innermost can be oriented:

pred(s(X)) -> X

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(GCD(x₁, x₂)) = x₁

POL(pred(x₁)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(IF(x₁, x₂, x₃)) = x₂

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 21

                 ↳Dependency Graph

Dependency Pair:

GCD(s(s(Y''')), s(s(0))) -> IF(true, s(s(Y''')), s(s(0)))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(true, s(X'''), s(s(s(Y')))) -> GCD(pred(pred(minus(X''', Y'))), s(s(s(Y'))))
GCD(s(s(s(Y'))), s(s(s(X')))) -> IF(le(X', Y'), s(s(s(Y'))), s(s(s(X'))))
IF(false, s(s(Y'')), s(Y0)) -> GCD(pred(minus(Y0, Y'')), s(s(Y'')))
GCD(s(s(0)), s(s(s(X')))) -> IF(false, s(s(0)), s(s(s(X'))))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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(false, s(s(Y'')), s(Y0)) -> GCD(pred(minus(Y0, Y'')), s(s(Y'')))

two new Dependency Pairs are created:

IF(false, s(s(s(Y'))), s(Y0')) -> GCD(pred(pred(minus(Y0', Y'))), s(s(s(Y'))))
IF(false, s(s(0)), s(Y0')) -> GCD(pred(Y0'), s(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 19

                 ↳Polynomial Ordering

Dependency Pairs:

IF(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(le(X', Y'), s(s(s(Y'))), s(s(s(X'))))
IF(true, s(X'''), s(s(s(Y')))) -> GCD(pred(pred(minus(X''', Y'))), s(s(s(Y'))))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) -> gcd(minus(Y, X), s(X))

Strategy:

innermost

The following dependency pairs can be strictly oriented:

IF(false, s(s(s(Y'))), s(Y0')) -> GCD(pred(pred(minus(Y0', Y'))), s(s(s(Y'))))
IF(true, s(X'''), s(s(s(Y')))) -> GCD(pred(pred(minus(X''', Y'))), s(s(s(Y'))))

Additionally, the following usable rules for innermost can be oriented:

pred(s(X)) -> X
minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(GCD(x₁, x₂)) = x₁ + x₂

POL(false) = 0

POL(pred(x₁)) = x₁

POL(minus(x₁, x₂)) = x₁

POL(true) = 0

POL(s(x₁)) = 1 + x₁

POL(le(x₁, x₂)) = 0

POL(IF(x₁, x₂, x₃)) = x₂ + x₃

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 22

                 ↳Dependency Graph

Dependency Pair:

GCD(s(s(s(Y'))), s(s(s(X')))) -> IF(le(X', Y'), s(s(s(Y'))), s(s(s(X'))))

Rules:

minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X
pred(s(X)) -> X
le(s(X), s(Y)) -> le(X, Y)
le(s(X), 0) -> false
le(0, Y) -> true
gcd(0, Y) -> 0
gcd(s(X), 0) -> s(X)
gcd(s(X), s(Y)) -> if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) -> gcd(minus(X, Y), s(Y))
if(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:
0:00 minutes

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

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

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