Certification Problem
Input (TPDB TRS_Standard/SK90/4.52)
The rewrite relation of the following TRS is considered.
|
s(a) |
→ |
a |
(1) |
|
s(s(x)) |
→ |
x |
(2) |
|
s(f(x,y)) |
→ |
f(s(y),s(x)) |
(3) |
|
s(g(x,y)) |
→ |
g(s(x),s(y)) |
(4) |
|
f(x,a) |
→ |
x |
(5) |
|
f(a,y) |
→ |
y |
(6) |
|
f(g(x,y),g(u,v)) |
→ |
g(f(x,u),f(y,v)) |
(7) |
|
g(a,a) |
→ |
a |
(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(x,y),g(u,v)) |
→ |
f#(x,u) |
(9) |
|
s#(f(x,y)) |
→ |
s#(x) |
(10) |
|
s#(g(x,y)) |
→ |
s#(y) |
(11) |
|
s#(g(x,y)) |
→ |
g#(s(x),s(y)) |
(12) |
|
s#(f(x,y)) |
→ |
s#(y) |
(13) |
|
f#(g(x,y),g(u,v)) |
→ |
f#(y,v) |
(14) |
|
f#(g(x,y),g(u,v)) |
→ |
g#(f(x,u),f(y,v)) |
(15) |
|
s#(g(x,y)) |
→ |
s#(x) |
(16) |
|
s#(f(x,y)) |
→ |
f#(s(y),s(x)) |
(17) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.