Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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:

MIN(s(X), s(Y)) -> MIN(X, Y)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)
LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore, R contains three SCCs.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Usable Rules (Innermost)

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

MIN(s(X), s(Y)) -> MIN(X, Y)

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 6 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:

MIN(s(X), s(Y)) -> MIN(X, Y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

MIN(s(X), s(Y)) -> MIN(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:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 4 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

...

               →DP Problem 5

                 ↳Negative Polynomial Order

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

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

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

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

min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x₁, x₂) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( min(x₁, x₂) ) = x₁

This results in one new DP problem.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

...

               →DP Problem 6

                 ↳Dependency Graph

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

Rules:

min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳Usable Rules (Innermost)

Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 7

                 ↳Negative Polynomial Order

Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Rules:

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

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

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

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

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Used ordering:
Polynomial Order with Interpretation:

POL( LOG(x₁) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( quot(x₁, x₂) ) = x₁

POL( 0 ) = 0

POL( min(x₁, x₂) ) = x₁

This results in one new DP problem.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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