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

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:

LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), s(y)) -> MINUS(x, y)
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) -> LE(y, x)
IFMOD(true, x, y) -> MOD(minus(x, y), y)
IFMOD(true, x, y) -> MINUS(x, y)

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:

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)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, x, y) -> mod(minus(x, y), y)
ifmod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

As we are in the innermost case, we can delete all 10 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:

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
Usable Rules (Innermost)
→DP Problem 3
UsableRules

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, x, y) -> mod(minus(x, y), y)
ifmod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

As we are in the innermost case, we can delete all 10 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:

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
UsableRules
→DP Problem 3
Usable Rules (Innermost)

Dependency Pairs:

IFMOD(true, x, y) -> MOD(minus(x, y), y)
MOD(s(x), s(y)) -> IFMOD(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)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, x, y) -> mod(minus(x, y), y)
ifmod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 6
Narrowing Transformation

Dependency Pairs:

IFMOD(true, x, y) -> MOD(minus(x, y), y)
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))

Rules:

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

Strategy:

innermost

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

IFMOD(true, x, y) -> MOD(minus(x, y), y)
two new Dependency Pairs are created:

IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
IFMOD(true, x'', 0) -> MOD(x'', 0)

The transformation is resulting in one new DP problem:

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 7
Narrowing Transformation

Dependency Pairs:

IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))

Rules:

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

Strategy:

innermost

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

MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))
three new Dependency Pairs are created:

MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))
MOD(s(0), s(s(x''))) -> IFMOD(false, s(0), s(s(x'')))

The transformation is resulting in one new DP problem:

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 8
Instantiation Transformation

Dependency Pairs:

MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))
MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))
IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))

Rules:

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

Strategy:

innermost

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

IFMOD(true, s(x''), s(y'')) -> MOD(minus(x'', y''), s(y''))
two new Dependency Pairs are created:

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))
IFMOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))

The transformation is resulting in two new DP problems:

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 9
Rewriting Transformation

Dependency Pairs:

IFMOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))

Rules:

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

Strategy:

innermost

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

IFMOD(true, s(x''''), s(0)) -> MOD(minus(x'''', 0), s(0))
one new Dependency Pair is created:

IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))

The transformation is resulting in one new DP problem:

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 11
Usable Rules (Innermost)

Dependency Pairs:

IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 13
Size-Change Principle

Dependency Pairs:

IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))
MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. IFMOD(true, s(x''''), s(0)) -> MOD(x'''', s(0))
2. MOD(s(x'), s(0)) -> IFMOD(true, s(x'), s(0))
and get the following Size-Change Graph(s):
{1} , {1}
2>1
3=2
{2} , {2}
1=2
2=3

which lead(s) to this/these maximal multigraph(s):
{1} , {2}
2>2
3=3
{2} , {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
UsableRules
...
→DP Problem 10
Rewriting Transformation

Dependency Pairs:

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))
MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

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

Strategy:

innermost

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

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(s(y''''), s(x'''')), s(s(x'''')))
one new Dependency Pair is created:

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(y'''', x''''), s(s(x'''')))

The transformation is resulting in one new DP problem:

R
OC
→TRS2
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
...
→DP Problem 12
Negative Polynomial Order

Dependency Pairs:

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(y'''', x''''), s(s(x'''')))
MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

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

Strategy:

innermost

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

IFMOD(true, s(s(y'''')), s(s(x''''))) -> MOD(minus(y'''', x''''), s(s(x'''')))

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( IFMOD(x1, ..., x3) ) = x2

POL( s(x1) ) = x1 + 1

POL( MOD(x1, x2) ) = x1

POL( minus(x1, x2) ) = x1

POL( le(x1, x2) ) = 0

POL( true ) = 0

POL( false ) = 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 14
Dependency Graph

Dependency Pair:

MOD(s(s(y'')), s(s(x''))) -> IFMOD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Termination of R successfully shown.
Duration:
0:07 minutes