Certification Problem
Input (TPDB TRS_Standard/SK90/2.10)
The rewrite relation of the following TRS is considered.
|
minus(0) |
→ |
0 |
(1) |
|
+(x,0) |
→ |
x |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(minus(1),1) |
→ |
0 |
(4) |
|
minus(minus(x)) |
→ |
x |
(5) |
|
+(x,minus(y)) |
→ |
minus(+(minus(x),y)) |
(6) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(7) |
|
+(minus(+(x,1)),1) |
→ |
minus(x) |
(8) |
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,minus(y)) |
→ |
+#(minus(x),y) |
(9) |
|
+#(minus(+(x,1)),1) |
→ |
minus#(x) |
(10) |
|
+#(x,+(y,z)) |
→ |
+#(x,y) |
(11) |
|
+#(x,minus(y)) |
→ |
minus#(x) |
(12) |
|
+#(x,minus(y)) |
→ |
minus#(+(minus(x),y)) |
(13) |
|
+#(x,+(y,z)) |
→ |
+#(+(x,y),z) |
(14) |
1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.