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

Termination of R to be shown.

`   R`
`     ↳Overlay and local confluence Check`

The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), 0, z, u) -> MINUS(z, s(x))
F(s(x), s(y), z, u) -> IF(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) -> LE(x, y)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)
F(s(x), s(y), z, u) -> MINUS(y, x)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Furthermore, R contains three SCCs.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳Usable Rules (Innermost)`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

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

Rules:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 non-usable-rules.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`             ...`
`               →DP Problem 4`
`                 ↳Size-Change Principle`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. MINUS(s(x), s(y)) -> MINUS(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳Usable Rules (Innermost)`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

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

Rules:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 non-usable-rules.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`             ...`
`               →DP Problem 5`
`                 ↳Size-Change Principle`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. LE(s(x), s(y)) -> LE(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳Usable Rules (Innermost)`

Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

Rules:

minus(0, y) -> 0
minus(s(x), 0) -> s(x)
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, minus(z, s(x)), u)
f(s(x), s(y), z, u) -> if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

As we are in the innermost case, we can delete all 11 non-usable-rules.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`
`             ...`
`               →DP Problem 6`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(0, y) -> 0
minus(s(x), 0) -> s(x)

Strategy:

innermost

The following Dependency Pairs can be strictly oriented using the given order.

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

minus(s(x), s(y)) -> minus(x, y)
minus(0, y) -> 0
minus(s(x), 0) -> s(x)

Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x4) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

POL( 0 ) = 0

This results in one new DP problem.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`
`             ...`
`               →DP Problem 7`
`                 ↳Negative Polynomial Order`

Dependency Pair:

F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(0, y) -> 0
minus(s(x), 0) -> s(x)

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

minus(s(x), s(y)) -> minus(x, y)
minus(0, y) -> 0
minus(s(x), 0) -> s(x)

Used ordering:
Polynomial Order with Interpretation:

POL( F(x1, ..., x4) ) = x2

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

POL( 0 ) = 0

This results in one new DP problem.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(0, y) -> 0
minus(s(x), 0) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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