Term Rewriting System R:
[x]
+(1, x) -> +(+(0, 1), x)
+(0, x) -> x
Innermost Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
+'(1, x) -> +'(+(0, 1), x)
+'(1, x) -> +'(0, 1)
Furthermore, R contains one SCC.
R
↳DPs
→DP Problem 1
↳Usable Rules (Innermost)
Dependency Pair:
+'(1, x) -> +'(+(0, 1), x)
Rules:
+(1, x) -> +(+(0, 1), x)
+(0, x) -> x
Strategy:
innermost
As we are in the innermost case, we can delete all 1 non-usable-rules.
R
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳Rewriting Transformation
Dependency Pair:
+'(1, x) -> +'(+(0, 1), x)
Rule:
+(0, x) -> x
Strategy:
innermost
On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule
+'(1, x) -> +'(+(0, 1), x)
one new Dependency Pair
is created:
+'(1, x) -> +'(1, x)
The transformation is resulting in one new DP problem:
R
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳Rw
...
→DP Problem 3
↳Usable Rules (Innermost)
Dependency Pair:
+'(1, x) -> +'(1, x)
Rule:
+(0, x) -> x
Strategy:
innermost
As we are in the innermost case, we can delete all 1 non-usable-rules.
R
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳Rw
...
→DP Problem 4
↳A-Transformation
Dependency Pair:
+'(1, x) -> +'(1, x)
Rule:
none
Strategy:
innermost
We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.
R
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳Rw
...
→DP Problem 5
↳Non Termination
Dependency Pair:
1'(x) -> 1'(x)
Rule:
none
Strategy:
innermost
Found an infinite P-chain over R:
P =
1'(x) -> 1'(x)
R = none
s = 1'(x')
evaluates to t =1'(x')
Thus, s starts an infinite chain.
Innermost Non-Termination of R could be shown.
Duration:
0:03 minutes