Term Rewriting System R:
[x, y, z, u, v]
if(if(x, y, z), u, v) -> if(x, if(y, u, v), if(z, u, v))
Innermost Termination of R to be shown.
R
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
IF(if(x, y, z), u, v) -> IF(x, if(y, u, v), if(z, u, v))
IF(if(x, y, z), u, v) -> IF(y, u, v)
IF(if(x, y, z), u, v) -> IF(z, u, v)
Furthermore, R contains one SCC.
R
↳DPs
→DP Problem 1
↳Polynomial Ordering
Dependency Pairs:
IF(if(x, y, z), u, v) -> IF(z, u, v)
IF(if(x, y, z), u, v) -> IF(y, u, v)
Rule:
if(if(x, y, z), u, v) -> if(x, if(y, u, v), if(z, u, v))
Strategy:
innermost
The following dependency pairs can be strictly oriented:
IF(if(x, y, z), u, v) -> IF(z, u, v)
IF(if(x, y, z), u, v) -> IF(y, u, v)
There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
POL(if(x1, x2, x3)) | = 1 + x2 + x3 |
POL(IF(x1, x2, x3)) | = x1 |
resulting in one new DP problem.
R
↳DPs
→DP Problem 1
↳Polo
→DP Problem 2
↳Dependency Graph
Dependency Pair:
Rule:
if(if(x, y, z), u, v) -> if(x, if(y, u, v), if(z, u, v))
Strategy:
innermost
Using the Dependency Graph resulted in no new DP problems.
Innermost Termination of R successfully shown.
Duration:
0:00 minutes