Certification Problem
Input (TPDB TRS_Standard/SK90/4.09)
The rewrite relation of the following TRS is considered.
|
+(x,0) |
→ |
x |
(1) |
|
+(x,i(x)) |
→ |
0 |
(2) |
|
+(+(x,y),z) |
→ |
+(x,+(y,z)) |
(3) |
|
*(x,+(y,z)) |
→ |
+(*(x,y),*(x,z)) |
(4) |
|
*(+(x,y),z) |
→ |
+(*(x,z),*(y,z)) |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
*#(+(x,y),z) |
→ |
*#(x,z) |
(6) |
|
*#(x,+(y,z)) |
→ |
+#(*(x,y),*(x,z)) |
(7) |
|
*#(x,+(y,z)) |
→ |
*#(x,y) |
(8) |
|
+#(+(x,y),z) |
→ |
+#(y,z) |
(9) |
|
*#(+(x,y),z) |
→ |
*#(y,z) |
(10) |
|
*#(+(x,y),z) |
→ |
+#(*(x,z),*(y,z)) |
(11) |
|
*#(x,+(y,z)) |
→ |
*#(x,z) |
(12) |
|
+#(+(x,y),z) |
→ |
+#(x,+(y,z)) |
(13) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.