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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

P(p(s(x))) -> P(x)
LE(p(s(x)), x) -> LE(x, x)
LE(s(x), s(y)) -> LE(x, y)
MINUS(x, y) -> IF(le(x, y), x, y)
MINUS(x, y) -> LE(x, y)
IF(false, x, y) -> MINUS(p(x), y)
IF(false, x, y) -> P(x)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

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 3`
`             ↳Forward Instantiation Transformation`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

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 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 4`
`                 ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

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

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

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)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`           →DP Problem 3`
`             ↳FwdInst`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Nar`

Dependency Pair:

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Narrowing Transformation`

Dependency Pairs:

IF(false, x, y) -> MINUS(p(x), y)
MINUS(x, y) -> IF(le(x, y), x, y)

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

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

MINUS(x, y) -> IF(le(x, y), x, y)
three new Dependency Pairs are created:

MINUS(0, y'') -> IF(true, 0, y'')
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Narrowing Transformation`

Dependency Pairs:

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, x, y) -> MINUS(p(x), y)

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

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

IF(false, x, y) -> MINUS(p(x), y)
two new Dependency Pairs are created:

IF(false, 0, y) -> MINUS(s(s(0)), y)
IF(false, s(x''), y) -> MINUS(x'', y)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Narrowing Transformation`

Dependency Pairs:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, s(x''), y) -> MINUS(x'', y)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

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

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))
three new Dependency Pairs are created:

MINUS(s(0), s(y''')) -> IF(true, s(0), s(y'''))
MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))
MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Instantiation Transformation`

Dependency Pairs:

MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))
MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))
IF(false, s(x''), y) -> MINUS(x'', y)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

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

IF(false, s(x''), y) -> MINUS(x'', y)
three new Dependency Pairs are created:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
IF(false, s(s(x'''')), s(0)) -> MINUS(s(x''''), s(0))
IF(false, s(s(x'''')), s(s(y'''))) -> MINUS(s(x''''), s(s(y''')))

The transformation is resulting in three new DP problems:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

IF(false, s(s(x'''')), s(s(y'''))) -> MINUS(s(x''''), s(s(y''')))
MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(false) =  0 POL(MINUS(x1, x2)) =  1 + x1 + x2 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(le) =  0 POL(IF(x1, x2, x3)) =  x1 + x2 + x3

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
IF(x1, x2, x3) -> IF(x1, x2, x3)
s(x1) -> s(x1)
le(x1, x2) -> le

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 12`
`                 ↳Dependency Graph`

Dependency Pair:

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

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 10`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

IF(false, s(s(x'''')), s(0)) -> MINUS(s(x''''), s(0))
MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS(s(s(x')), s(0)) -> IF(false, s(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(false) =  0 POL(MINUS(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  1 + x1 POL(IF(x1, x2, x3)) =  x1 + x2 + x3

resulting in one new DP problem.
Used Argument Filtering System:
MINUS(x1, x2) -> MINUS(x1, x2)
IF(x1, x2, x3) -> IF(x1, x2, x3)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 13`
`                 ↳Dependency Graph`

Dependency Pair:

IF(false, s(s(x'''')), s(0)) -> MINUS(s(x''''), s(0))

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 11`
`                 ↳Argument Filtering and Ordering`

Dependency Pairs:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

IF(false, s(x''''), 0) -> MINUS(x'''', 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(false) =  0 POL(MINUS(x1, x2)) =  x1 + x2 POL(s(x1)) =  1 + x1 POL(IF(x1, x2, x3)) =  x1 + x2 + x3

resulting in one new DP problem.
Used Argument Filtering System:
IF(x1, x2, x3) -> IF(x1, x2, x3)
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳FwdInst`
`       →DP Problem 2`
`         ↳Nar`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 14`
`                 ↳Dependency Graph`

Dependency Pair:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)

Rules:

p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes