Certification Problem
Input (TPDB TRS_Standard/Transformed_CSR_04/ExConc_Zan97_GM)
The rewrite relation of the following TRS is considered.
a__f(X) |
→ |
g(h(f(X))) |
(1) |
mark(f(X)) |
→ |
a__f(mark(X)) |
(2) |
mark(g(X)) |
→ |
g(X) |
(3) |
mark(h(X)) |
→ |
h(mark(X)) |
(4) |
a__f(X) |
→ |
f(X) |
(5) |
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
a__f(X) |
→ |
f(h(g(X))) |
(6) |
f(mark(X)) |
→ |
mark(a__f(X)) |
(7) |
g(mark(X)) |
→ |
g(X) |
(8) |
h(mark(X)) |
→ |
mark(h(X)) |
(9) |
a__f(X) |
→ |
f(X) |
(5) |
1.1 Rule Removal
Using the
linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1
over the naturals
[mark(x1)] |
= |
· x1 +
|
[f(x1)] |
= |
· x1 +
|
[g(x1)] |
= |
· x1 +
|
[a__f(x1)] |
= |
· x1 +
|
[h(x1)] |
= |
· x1 +
|
all of the following rules can be deleted.
a__f(X) |
→ |
f(h(g(X))) |
(6) |
g(mark(X)) |
→ |
g(X) |
(8) |
a__f(X) |
→ |
f(X) |
(5) |
1.1.1 Rule Removal
Using the
Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(mark) |
= |
0 |
|
weight(mark) |
= |
2 |
|
|
|
prec(h) |
= |
1 |
|
weight(h) |
= |
2 |
|
|
|
prec(f) |
= |
3 |
|
weight(f) |
= |
2 |
|
|
|
prec(a__f) |
= |
2 |
|
weight(a__f) |
= |
2 |
|
|
|
all of the following rules can be deleted.
f(mark(X)) |
→ |
mark(a__f(X)) |
(7) |
h(mark(X)) |
→ |
mark(h(X)) |
(9) |
1.1.1.1 R is empty
There are no rules in the TRS. Hence, it is terminating.