Certification Problem
Input (TPDB SRS_Standard/Zantema_04/z122)
The rewrite relation of the following TRS is considered.
a(a(x1)) |
→ |
b(c(x1)) |
(1) |
b(b(x1)) |
→ |
c(d(x1)) |
(2) |
c(c(x1)) |
→ |
d(d(d(x1))) |
(3) |
d(c(x1)) |
→ |
b(f(x1)) |
(4) |
d(d(d(x1))) |
→ |
a(c(x1)) |
(5) |
f(f(x1)) |
→ |
f(b(x1)) |
(6) |
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
[f(x1)] |
= |
x1 +
|
[d(x1)] |
= |
x1 +
|
[c(x1)] |
= |
x1 +
|
[b(x1)] |
= |
x1 +
|
[a(x1)] |
= |
x1 +
|
all of the following rules can be deleted.
a(a(x1)) |
→ |
b(c(x1)) |
(1) |
c(c(x1)) |
→ |
d(d(d(x1))) |
(3) |
d(d(d(x1))) |
→ |
a(c(x1)) |
(5) |
1.1 String Reversal
Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(b(x1)) |
→ |
d(c(x1)) |
(7) |
c(d(x1)) |
→ |
f(b(x1)) |
(8) |
f(f(x1)) |
→ |
b(f(x1)) |
(9) |
1.1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
f#(f(x1)) |
→ |
b#(f(x1)) |
(10) |
c#(d(x1)) |
→ |
f#(b(x1)) |
(11) |
c#(d(x1)) |
→ |
b#(x1) |
(12) |
b#(b(x1)) |
→ |
c#(x1) |
(13) |
1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[f(x1)] |
= |
x1 +
|
[d(x1)] |
= |
x1 +
|
[c(x1)] |
= |
x1 +
|
[b(x1)] |
= |
x1 +
|
[f#(x1)] |
= |
x1 +
|
[c#(x1)] |
= |
x1 +
|
[b#(x1)] |
= |
x1 +
|
together with the usable
rules
b(b(x1)) |
→ |
d(c(x1)) |
(7) |
c(d(x1)) |
→ |
f(b(x1)) |
(8) |
f(f(x1)) |
→ |
b(f(x1)) |
(9) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pairs
f#(f(x1)) |
→ |
b#(f(x1)) |
(10) |
c#(d(x1)) |
→ |
f#(b(x1)) |
(11) |
c#(d(x1)) |
→ |
b#(x1) |
(12) |
b#(b(x1)) |
→ |
c#(x1) |
(13) |
and
no rules
could be deleted.
1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.