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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
-(#, x) -> #
-(x, #) -> x
-(0(x), 0(y)) -> 0(-(x, y))
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) -> 1(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
not(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 0(y)) -> ge(x, y)
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)

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)
+'(1(x), 1(y)) -> 0'(+(+(x, y), 1(#)))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
-'(0(x), 0(y)) -> 0'(-(x, y))
-'(0(x), 0(y)) -> -'(x, y)
-'(0(x), 1(y)) -> -'(-(x, y), 1(#))
-'(0(x), 1(y)) -> -'(x, y)
-'(1(x), 0(y)) -> -'(x, y)
-'(1(x), 1(y)) -> 0'(-(x, y))
-'(1(x), 1(y)) -> -'(x, y)
GE(0(x), 0(y)) -> GE(x, y)
GE(0(x), 1(y)) -> NOT(ge(y, x))
GE(0(x), 1(y)) -> GE(y, x)
GE(1(x), 0(y)) -> GE(x, y)
GE(1(x), 1(y)) -> GE(x, y)
GE(#, 0(x)) -> GE(#, x)
LOG(x) -> -'(log'(x), 1(#))
LOG(x) -> LOG'(x)
LOG'(1(x)) -> +'(log'(x), 1(#))
LOG'(1(x)) -> LOG'(x)
LOG'(0(x)) -> IF(ge(x, 1(#)), +(log'(x), 1(#)), #)
LOG'(0(x)) -> GE(x, 1(#))
LOG'(0(x)) -> +'(log'(x), 1(#))
LOG'(0(x)) -> LOG'(x)

Furthermore, R contains five SCCs.


   R
DPs
       →DP Problem 1
Narrowing Transformation
       →DP Problem 2
Remaining
       →DP Problem 3
Remaining
       →DP Problem 4
Remaining
       →DP Problem 5
Remaining


Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(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))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
-(#, x) -> #
-(x, #) -> x
-(0(x), 0(y)) -> 0(-(x, y))
-(0(x), 1(y)) -> 1(-(-(x, y), 1(#)))
-(1(x), 0(y)) -> 1(-(x, y))
-(1(x), 1(y)) -> 0(-(x, y))
not(true) -> false
not(false) -> true
if(true, x, y) -> x
if(false, x, y) -> y
ge(0(x), 0(y)) -> ge(x, y)
ge(0(x), 1(y)) -> not(ge(y, x))
ge(1(x), 0(y)) -> ge(x, y)
ge(1(x), 1(y)) -> ge(x, y)
ge(x, #) -> true
ge(#, 0(x)) -> ge(#, x)
ge(#, 1(x)) -> false
log(x) -> -(log'(x), 1(#))
log'(#) -> #
log'(1(x)) -> +(log'(x), 1(#))
log'(0(x)) -> if(ge(x, 1(#)), +(log'(x), 1(#)), #)


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))
six new Dependency Pairs are created:

+'(+(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, 1(x'')), 1(y'')) -> +'(x, 0(+(+(x'', y''), 1(#))))
+'(+(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)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
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)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
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)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
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)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
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)
       →DP Problem 4
Remaining Obligation(s)
       →DP Problem 5
Remaining Obligation(s)




The following remains to be proven:

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