Certification Problem
Input (TPDB TRS_Standard/Various_04/12)
The rewrite relation of the following TRS is considered.
|
O(0) |
→ |
0 |
(1) |
|
+(0,x) |
→ |
x |
(2) |
|
+(x,0) |
→ |
x |
(3) |
|
+(O(x),O(y)) |
→ |
O(+(x,y)) |
(4) |
|
+(O(x),I(y)) |
→ |
I(+(x,y)) |
(5) |
|
+(I(x),O(y)) |
→ |
I(+(x,y)) |
(6) |
|
+(I(x),I(y)) |
→ |
O(+(+(x,y),I(0))) |
(7) |
|
*(0,x) |
→ |
0 |
(8) |
|
*(x,0) |
→ |
0 |
(9) |
|
*(O(x),y) |
→ |
O(*(x,y)) |
(10) |
|
*(I(x),y) |
→ |
+(O(*(x,y)),y) |
(11) |
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.
|
*#(I(x),y) |
→ |
+#(O(*(x,y)),y) |
(12) |
|
+#(I(x),I(y)) |
→ |
+#(+(x,y),I(0)) |
(13) |
|
*#(O(x),y) |
→ |
*#(x,y) |
(14) |
|
+#(O(x),O(y)) |
→ |
O#(+(x,y)) |
(15) |
|
+#(O(x),O(y)) |
→ |
+#(x,y) |
(16) |
|
+#(I(x),I(y)) |
→ |
+#(x,y) |
(17) |
|
+#(O(x),I(y)) |
→ |
+#(x,y) |
(18) |
|
+#(I(x),I(y)) |
→ |
O#(+(+(x,y),I(0))) |
(19) |
|
*#(I(x),y) |
→ |
O#(*(x,y)) |
(20) |
|
+#(I(x),O(y)) |
→ |
+#(x,y) |
(21) |
|
*#(O(x),y) |
→ |
O#(*(x,y)) |
(22) |
|
*#(I(x),y) |
→ |
*#(x,y) |
(23) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.