Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/Ex4_7_15_Bor03_GM)
The rewrite relation of the following TRS is considered.
a__f(0) |
→ |
cons(0,f(s(0))) |
(1) |
a__f(s(0)) |
→ |
a__f(a__p(s(0))) |
(2) |
a__p(s(0)) |
→ |
0 |
(3) |
mark(f(X)) |
→ |
a__f(mark(X)) |
(4) |
mark(p(X)) |
→ |
a__p(mark(X)) |
(5) |
mark(0) |
→ |
0 |
(6) |
mark(cons(X1,X2)) |
→ |
cons(mark(X1),X2) |
(7) |
mark(s(X)) |
→ |
s(mark(X)) |
(8) |
a__f(X) |
→ |
f(X) |
(9) |
a__p(X) |
→ |
p(X) |
(10) |
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.
a__f#(s(0)) |
→ |
a__p#(s(0)) |
(11) |
mark#(s(X)) |
→ |
mark#(X) |
(12) |
mark#(f(X)) |
→ |
a__f#(mark(X)) |
(13) |
mark#(p(X)) |
→ |
mark#(X) |
(14) |
mark#(cons(X1,X2)) |
→ |
mark#(X1) |
(15) |
a__f#(s(0)) |
→ |
a__f#(a__p(s(0))) |
(16) |
mark#(f(X)) |
→ |
mark#(X) |
(17) |
mark#(p(X)) |
→ |
a__p#(mark(X)) |
(18) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.