Term Rewriting System R:
[x, y, z]
0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(j(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(j(x), j(y)) -> +'(x, y)
+'(1(x), j(y)) -> 0'(+(x, y))
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), 1(y)) -> 0'(+(x, y))
+'(j(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
OPP(0(x)) -> 0'(opp(x))
OPP(0(x)) -> OPP(x)
OPP(1(x)) -> OPP(x)
OPP(j(x)) -> OPP(x)
-'(x, y) -> +'(x, opp(y))
-'(x, y) -> OPP(y)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(j(x), y) -> -'(0(*(x, y)), y)
*'(j(x), y) -> 0'(*(x, y))
*'(j(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(j(x), 1(y)) -> +'(x, y)
+'(1(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(x, y)
+'(j(x), j(y)) -> +'(+(x, y), j(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(j(x), 0(y)) -> +'(x, y)
+'(0(x), j(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)


Rules:


0(#) -> #
+(#, x) -> x
+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(0(x), j(y)) -> j(+(x, y))
+(j(x), 0(y)) -> j(+(x, y))
+(1(x), 1(y)) -> j(+(+(x, y), 1(#)))
+(j(x), j(y)) -> 1(+(+(x, y), j(#)))
+(1(x), j(y)) -> 0(+(x, y))
+(j(x), 1(y)) -> 0(+(x, y))
+(+(x, y), z) -> +(x, +(y, z))
opp(#) -> #
opp(0(x)) -> 0(opp(x))
opp(1(x)) -> j(opp(x))
opp(j(x)) -> 1(opp(x))
-(x, y) -> +(x, opp(y))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(j(x), y) -> -(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))


Strategy:

innermost




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

+'(+(x, y), z) -> +'(x, +(y, z))
12 new Dependency Pairs are created:

+'(+(x, #), z') -> +'(x, z')
+'(+(x, y'), #) -> +'(x, y')
+'(+(x, 0(x'')), 0(y'')) -> +'(x, 0(+(x'', y'')))
+'(+(x, 0(x'')), 1(y'')) -> +'(x, 1(+(x'', y'')))
+'(+(x, 1(x'')), 0(y'')) -> +'(x, 1(+(x'', y'')))
+'(+(x, 0(x'')), j(y'')) -> +'(x, j(+(x'', y'')))
+'(+(x, j(x'')), 0(y'')) -> +'(x, j(+(x'', y'')))
+'(+(x, 1(x'')), 1(y'')) -> +'(x, j(+(+(x'', y''), 1(#))))
+'(+(x, j(x'')), j(y'')) -> +'(x, 1(+(+(x'', y''), j(#))))
+'(+(x, 1(x'')), j(y'')) -> +'(x, 0(+(x'', y'')))
+'(+(x, j(x'')), 1(y'')) -> +'(x, 0(+(x'', y'')))
+'(+(x, +(x'', y'')), z'') -> +'(x, +(x'', +(y'', z'')))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
Nar
       →DP Problem 2
Remaining Obligation(s)
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:05 minutes