Term Rewriting System R:
[X, Y]
*(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
*(X, 1) -> X
*(X, 0) -> X
*(X, 0) -> 0
Innermost Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))
*'(X, +(Y, 1)) -> *'(1, 0)
Furthermore, R contains one SCC.
R
↳DPs
→DP Problem 1
↳Usable Rules (Innermost)
Dependency Pair:
*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))
Rules:
*(X, +(Y, 1)) -> +(*(X, +(Y, *(1, 0))), X)
*(X, 1) -> X
*(X, 0) -> X
*(X, 0) -> 0
Strategy:
innermost
As we are in the innermost case, we can delete all 2 non-usable-rules.
R
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳Non Termination
Dependency Pair:
*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))
Rules:
*(X, 0) -> 0
*(X, 0) -> X
Strategy:
innermost
Found an infinite P-chain over R:
P =
*'(X, +(Y, 1)) -> *'(X, +(Y, *(1, 0)))
R =
*(X, 0) -> 0
*(X, 0) -> X
s = *'(X', +(Y', *(1, 0)))
evaluates to t =*'(X', +(Y', *(1, 0)))
Thus, s starts an infinite chain.
Innermost Non-Termination of R could be shown.
Duration:
0:00 minutes