Term Rewriting System R:
[x, y]
O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(O(x), O(y)) -> O'(+(x, y))
+'(O(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(I(x), O(y)) -> +'(x, y)
+'(I(x), I(y)) -> O'(+(+(x, y), I(0)))
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), I(y)) -> +'(x, y)
*'(O(x), y) -> O'(*(x, y))
*'(O(x), y) -> *'(x, y)
*'(I(x), y) -> +'(O(*(x, y)), y)
*'(I(x), y) -> O'(*(x, y))
*'(I(x), y) -> *'(x, y)

Furthermore, R contains two SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
Remaining


Dependency Pairs:

+'(I(x), I(y)) -> +'(x, y)
+'(I(x), I(y)) -> +'(+(x, y), I(0))
+'(I(x), O(y)) -> +'(x, y)
+'(O(x), I(y)) -> +'(x, y)
+'(O(x), O(y)) -> +'(x, y)


Rules:


O(0) -> 0
+(0, x) -> x
+(x, 0) -> x
+(O(x), O(y)) -> O(+(x, y))
+(O(x), I(y)) -> I(+(x, y))
+(I(x), O(y)) -> I(+(x, y))
+(I(x), I(y)) -> O(+(+(x, y), I(0)))
*(0, x) -> 0
*(x, 0) -> 0
*(O(x), y) -> O(*(x, y))
*(I(x), y) -> +(O(*(x, y)), y)


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

+'(I(x), I(y)) -> +'(+(x, y), I(0))
six new Dependency Pairs are created:

+'(I(0), I(y')) -> +'(y', I(0))
+'(I(x''), I(0)) -> +'(x'', I(0))
+'(I(O(x'')), I(O(y''))) -> +'(O(+(x'', y'')), I(0))
+'(I(O(x'')), I(I(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(I(x'')), I(O(y''))) -> +'(I(+(x'', y'')), I(0))
+'(I(I(x'')), I(I(y''))) -> +'(O(+(+(x'', y''), I(0))), I(0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:06 minutes