YES
The TRS could be proven terminating. The proof took 2353 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (750ms).
| Problem 2 was processed with processor PolynomialLinearRange4iUR (792ms).
| | Problem 5 was processed with processor PolynomialLinearRange4iUR (365ms).
| | | Problem 6 was processed with processor PolynomialLinearRange4iUR (157ms).
| Problem 3 was processed with processor SubtermCriterion (1ms).
| Problem 4 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
+#(j(x), 1(y)) | → | 0#(+(x, y)) | | +#(0(x), j(y)) | → | +#(x, y) |
+#(j(x), j(y)) | → | +#(x, y) | | opp#(0(x)) | → | opp#(x) |
+#(0(x), 0(y)) | → | +#(x, y) | | *#(*(x, y), z) | → | *#(x, *(y, z)) |
+#(+(x, y), z) | → | +#(y, z) | | *#(j(x), y) | → | -#(0(*(x, y)), y) |
*#(0(x), y) | → | 0#(*(x, y)) | | +#(j(x), 1(y)) | → | +#(x, y) |
+#(1(x), j(y)) | → | 0#(+(x, y)) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
+#(0(x), 0(y)) | → | 0#(+(x, y)) | | +#(1(x), j(y)) | → | +#(x, y) |
+#(j(x), j(y)) | → | +#(+(x, y), j(#)) | | *#(j(x), y) | → | 0#(*(x, y)) |
*#(0(x), y) | → | *#(x, y) | | +#(1(x), 0(y)) | → | +#(x, y) |
opp#(0(x)) | → | 0#(opp(x)) | | -#(x, y) | → | +#(x, opp(y)) |
+#(0(x), 1(y)) | → | +#(x, y) | | +#(j(x), 0(y)) | → | +#(x, y) |
*#(*(x, y), z) | → | *#(y, z) | | opp#(1(x)) | → | opp#(x) |
+#(1(x), 1(y)) | → | +#(+(x, y), 1(#)) | | -#(x, y) | → | opp#(y) |
*#(1(x), y) | → | +#(0(*(x, y)), y) | | opp#(j(x)) | → | opp#(x) |
*#(1(x), y) | → | *#(x, y) | | *#(1(x), y) | → | 0#(*(x, y)) |
*#(j(x), y) | → | *#(x, y) | | +#(1(x), 1(y)) | → | +#(x, y) |
Rewrite 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)) |
Original Signature
Termination of terms over the following signature is verified: #, 1, 0, opp, *, +, j, -
Strategy
The following SCCs where found
*#(1(x), y) → *#(x, y) | *#(*(x, y), z) → *#(y, z) |
*#(*(x, y), z) → *#(x, *(y, z)) | *#(j(x), y) → *#(x, y) |
*#(0(x), y) → *#(x, y) |
+#(0(x), j(y)) → +#(x, y) | +#(j(x), j(y)) → +#(x, y) |
+#(0(x), 1(y)) → +#(x, y) | +#(j(x), 0(y)) → +#(x, y) |
+#(0(x), 0(y)) → +#(x, y) | +#(+(x, y), z) → +#(y, z) |
+#(j(x), 1(y)) → +#(x, y) | +#(1(x), 1(y)) → +#(+(x, y), 1(#)) |
+#(+(x, y), z) → +#(x, +(y, z)) | +#(1(x), j(y)) → +#(x, y) |
+#(j(x), j(y)) → +#(+(x, y), j(#)) | +#(1(x), 1(y)) → +#(x, y) |
+#(1(x), 0(y)) → +#(x, y) |
opp#(j(x)) → opp#(x) | opp#(0(x)) → opp#(x) |
opp#(1(x)) → opp#(x) |
Problem 2: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
+#(0(x), j(y)) | → | +#(x, y) | | +#(j(x), j(y)) | → | +#(x, y) |
+#(0(x), 1(y)) | → | +#(x, y) | | +#(j(x), 0(y)) | → | +#(x, y) |
+#(0(x), 0(y)) | → | +#(x, y) | | +#(+(x, y), z) | → | +#(y, z) |
+#(j(x), 1(y)) | → | +#(x, y) | | +#(1(x), 1(y)) | → | +#(+(x, y), 1(#)) |
+#(+(x, y), z) | → | +#(x, +(y, z)) | | +#(1(x), j(y)) | → | +#(x, y) |
+#(j(x), j(y)) | → | +#(+(x, y), j(#)) | | +#(1(x), 1(y)) | → | +#(x, y) |
+#(1(x), 0(y)) | → | +#(x, y) |
Rewrite 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)) |
Original Signature
Termination of terms over the following signature is verified: #, 1, 0, opp, *, +, j, -
Strategy
Polynomial Interpretation
- #: 0
- *(x,y): 0
- +(x,y): y + x
- +#(x,y): y + x + 1
- -(x,y): 0
- 0(x): x + 1
- 1(x): x + 1
- j(x): x + 1
- opp(x): 0
Improved Usable rules
+(+(x, y), z) | → | +(x, +(y, z)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
0(#) | → | # | | +(0(x), 1(y)) | → | 1(+(x, y)) |
+(j(x), 0(y)) | → | j(+(x, y)) | | +(0(x), j(y)) | → | j(+(x, y)) |
+(j(x), 1(y)) | → | 0(+(x, y)) | | +(#, x) | → | x |
+(0(x), 0(y)) | → | 0(+(x, y)) | | +(x, #) | → | x |
+(1(x), 1(y)) | → | j(+(+(x, y), 1(#))) | | +(j(x), j(y)) | → | 1(+(+(x, y), j(#))) |
+(1(x), j(y)) | → | 0(+(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
+#(0(x), j(y)) | → | +#(x, y) | | +#(j(x), j(y)) | → | +#(x, y) |
+#(0(x), 1(y)) | → | +#(x, y) | | +#(j(x), 0(y)) | → | +#(x, y) |
+#(0(x), 0(y)) | → | +#(x, y) | | +#(j(x), 1(y)) | → | +#(x, y) |
+#(1(x), 1(y)) | → | +#(+(x, y), 1(#)) | | +#(1(x), j(y)) | → | +#(x, y) |
+#(j(x), j(y)) | → | +#(+(x, y), j(#)) | | +#(1(x), 1(y)) | → | +#(x, y) |
+#(1(x), 0(y)) | → | +#(x, y) |
Problem 5: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
+#(+(x, y), z) | → | +#(y, z) | | +#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite 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)) |
Original Signature
Termination of terms over the following signature is verified: #, 1, opp, 0, *, +, j, -
Strategy
Polynomial Interpretation
- #: 1
- *(x,y): 0
- +(x,y): y + x + 1
- +#(x,y): y + x
- -(x,y): 0
- 0(x): 1
- 1(x): 0
- j(x): 0
- opp(x): 0
Improved Usable rules
+(+(x, y), z) | → | +(x, +(y, z)) | | +(1(x), 0(y)) | → | 1(+(x, y)) |
0(#) | → | # | | +(0(x), 1(y)) | → | 1(+(x, y)) |
+(j(x), 0(y)) | → | j(+(x, y)) | | +(0(x), j(y)) | → | j(+(x, y)) |
+(j(x), 1(y)) | → | 0(+(x, y)) | | +(#, x) | → | x |
+(0(x), 0(y)) | → | 0(+(x, y)) | | +(x, #) | → | x |
+(1(x), 1(y)) | → | j(+(+(x, y), 1(#))) | | +(j(x), j(y)) | → | 1(+(+(x, y), j(#))) |
+(1(x), j(y)) | → | 0(+(x, y)) |
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
+#(+(x, y), z) | → | +#(y, z) |
Problem 6: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
+#(+(x, y), z) | → | +#(x, +(y, z)) |
Rewrite 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)) |
Original Signature
Termination of terms over the following signature is verified: #, 1, 0, opp, *, +, j, -
Strategy
Polynomial Interpretation
- #: 2
- *(x,y): 0
- +(x,y): x + 1
- +#(x,y): 2x
- -(x,y): 0
- 0(x): 2
- 1(x): 1
- j(x): 1
- opp(x): 0
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
+#(+(x, y), z) | → | +#(x, +(y, z)) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
*#(1(x), y) | → | *#(x, y) | | *#(*(x, y), z) | → | *#(y, z) |
*#(*(x, y), z) | → | *#(x, *(y, z)) | | *#(j(x), y) | → | *#(x, y) |
*#(0(x), y) | → | *#(x, y) |
Rewrite 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)) |
Original Signature
Termination of terms over the following signature is verified: #, 1, 0, opp, *, +, j, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
*#(1(x), y) | → | *#(x, y) | | *#(*(x, y), z) | → | *#(y, z) |
*#(*(x, y), z) | → | *#(x, *(y, z)) | | *#(j(x), y) | → | *#(x, y) |
*#(0(x), y) | → | *#(x, y) |
Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
opp#(j(x)) | → | opp#(x) | | opp#(0(x)) | → | opp#(x) |
opp#(1(x)) | → | opp#(x) |
Rewrite 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)) |
Original Signature
Termination of terms over the following signature is verified: #, 1, 0, opp, *, +, j, -
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
opp#(j(x)) | → | opp#(x) | | opp#(0(x)) | → | opp#(x) |
opp#(1(x)) | → | opp#(x) |