Certification Problem
Input (TPDB TRS_Standard/SK90/4.43)
The rewrite relation of the following TRS is considered.
|
+(x,0) |
→ |
x |
(1) |
|
+(x,s(y)) |
→ |
s(+(x,y)) |
(2) |
|
+(0,y) |
→ |
y |
(3) |
|
+(s(x),y) |
→ |
s(+(x,y)) |
(4) |
|
+(x,+(y,z)) |
→ |
+(+(x,y),z) |
(5) |
|
f(g(f(x))) |
→ |
f(h(s(0),x)) |
(6) |
|
f(g(h(x,y))) |
→ |
f(h(s(x),y)) |
(7) |
|
f(h(x,h(y,z))) |
→ |
f(h(+(x,y),z)) |
(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.
|
f#(g(f(x))) |
→ |
f#(h(s(0),x)) |
(9) |
|
f#(h(x,h(y,z))) |
→ |
f#(h(+(x,y),z)) |
(10) |
|
f#(h(x,h(y,z))) |
→ |
+#(x,y) |
(11) |
|
+#(x,+(y,z)) |
→ |
+#(x,y) |
(12) |
|
f#(g(h(x,y))) |
→ |
f#(h(s(x),y)) |
(13) |
|
+#(x,s(y)) |
→ |
+#(x,y) |
(14) |
|
+#(s(x),y) |
→ |
+#(x,y) |
(15) |
|
+#(x,+(y,z)) |
→ |
+#(+(x,y),z) |
(16) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.