Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z076)
The rewrite relation of the following TRS is considered.
|
f(s(x1)) |
→ |
s(s(f(p(s(x1))))) |
(1) |
|
f(0(x1)) |
→ |
0(x1) |
(2) |
|
p(s(x1)) |
→ |
x1 |
(3) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by ttt2 @ termCOMP 2023)
1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
|
s(f(x1)) |
→ |
s(p(f(s(s(x1))))) |
(4) |
|
0(f(x1)) |
→ |
0(x1) |
(5) |
|
s(p(x1)) |
→ |
x1 |
(6) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
| [f(x1)] |
= |
· x1 +
|
| [0(x1)] |
= |
· x1 +
|
| [s(x1)] |
= |
· x1 +
|
| [p(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
s#(f(x1)) |
→ |
s#(x1) |
(7) |
|
s#(f(x1)) |
→ |
s#(s(x1)) |
(8) |
|
s#(f(x1)) |
→ |
s#(p(f(s(s(x1))))) |
(9) |
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.