Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z078)
The rewrite relation of the following TRS is considered.
|
f(0(x1)) |
→ |
s(0(x1)) |
(1) |
|
d(0(x1)) |
→ |
0(x1) |
(2) |
|
d(s(x1)) |
→ |
s(s(d(p(s(x1))))) |
(3) |
|
f(s(x1)) |
→ |
d(f(p(s(x1)))) |
(4) |
|
p(s(x1)) |
→ |
x1 |
(5) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by matchbox @ termCOMP 2023)
1 Rule Removal
Using the
matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [0(x1)] |
= |
x1 +
|
| [f(x1)] |
= |
x1 +
|
| [d(x1)] |
= |
x1 +
|
| [s(x1)] |
= |
x1 +
|
| [p(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
f#(s(x1)) |
→ |
f#(p(s(x1))) |
(6) |
|
f#(s(x1)) |
→ |
d#(f(p(s(x1)))) |
(7) |
|
f#(s(x1)) |
→ |
p#(s(x1)) |
(8) |
|
d#(s(x1)) |
→ |
d#(p(s(x1))) |
(9) |
|
d#(s(x1)) |
→ |
p#(s(x1)) |
(10) |
1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
| [0(x1)] |
= |
x1 +
|
| [f(x1)] |
= |
x1 +
|
| [d(x1)] |
= |
x1 +
|
| [s(x1)] |
= |
x1 +
|
| [p(x1)] |
= |
x1 +
|
| [f#(x1)] |
= |
x1 +
|
| [d#(x1)] |
= |
x1 +
|
| [p#(x1)] |
= |
x1 +
|
together with the usable
rules
|
d(0(x1)) |
→ |
0(x1) |
(2) |
|
d(s(x1)) |
→ |
s(s(d(p(s(x1))))) |
(3) |
|
f(s(x1)) |
→ |
d(f(p(s(x1)))) |
(4) |
|
p(s(x1)) |
→ |
x1 |
(5) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
|
f#(s(x1)) |
→ |
d#(f(p(s(x1)))) |
(7) |
|
f#(s(x1)) |
→ |
p#(s(x1)) |
(8) |
|
d#(s(x1)) |
→ |
p#(s(x1)) |
(10) |
and
no rules
could be deleted.
1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.