Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z]
f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

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:

F(X) -> F(g(X))
F(X) -> G(X)
G(s(X)) -> G(X)
SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Furthermore, R contains three SCCs.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳Usable Rules (Innermost)

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

G(s(X)) -> G(X)

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

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 4

                 ↳Size-Change Principle

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

G(s(X)) -> G(X)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

G(s(X)) -> G(X)

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1

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:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

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 5

                 ↳Size-Change Principle

           →DP Problem 3

             ↳UsableRules

Dependency Pair:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

SEL(s(X), cons(Y, Z)) -> SEL(X, Z)

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:

cons(x₁, x₂) -> cons(x₁, x₂)
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 Pair:

F(X) -> F(g(X))

Rules:

f(X) -> cons(X, f(g(X)))
g(0) -> s(0)
g(s(X)) -> s(s(g(X)))
sel(0, cons(X, Y)) -> X
sel(s(X), cons(Y, Z)) -> sel(X, Z)

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

     ↳OC

       →TRS2

         ↳DPs

           →DP Problem 1

             ↳UsableRules

           →DP Problem 2

             ↳UsableRules

           →DP Problem 3

             ↳UsableRules

...

               →DP Problem 6

                 ↳Non Termination

Dependency Pair:

F(X) -> F(g(X))

Rules:

g(s(X)) -> s(s(g(X)))
g(0) -> s(0)

Strategy:

innermost

Found an infinite P-chain over R:
P =

F(X) -> F(g(X))

R =

g(s(X)) -> s(s(g(X)))
g(0) -> s(0)

s = F(X)
evaluates to t =F(g(X))

Thus, s starts an infinite chain as s matches t.

Non-Termination of R could be shown.
Duration:
0:01 minutes