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.

`   R`
`     ↳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.

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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 w.r.t. to the implicit AFS 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.

`   R`
`     ↳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.

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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 w.r.t. to the implicit AFS 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.

`   R`
`     ↳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.

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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 w.r.t. to the implicit AFS 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.

`   R`
`     ↳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.

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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:

`   R`
`     ↳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 w.r.t. to the implicit AFS 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.

`   R`
`     ↳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.

`   R`
`     ↳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:

`   R`
`     ↳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 w.r.t. to the implicit AFS can be oriented:

pred(s(X)) -> X
minus(X, s(Y)) -> pred(minus(X, Y))
minus(X, 0) -> X

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.

`   R`
`     ↳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:
0:01 minutes