Termination Proof using AProVE

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

Termination of R to be shown.

     ↳Overlay and local confluence Check

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

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

Furthermore, R contains three SCCs.

     ↳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)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳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:

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

D_P: 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.

     ↳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)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳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:

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

D_P: 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.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳Usable Rules (Innermost)

Dependency Pairs:

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IF_MOD(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)) -> if_mod(le(y, x), s(x), s(y))
if_mod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

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

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 6

                 ↳Negative Polynomial Order

Dependency Pairs:

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IF_MOD(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

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

IF_MOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))

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( IF_MOD(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.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pair:

MOD(s(x), s(y)) -> IF_MOD(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

Using the Dependency Graph resulted in no new DP problems.

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