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)] =
1 0 1
0 1 0
0 1 1
· x1 +
1 0 0
0 0 0
1 0 0
[f(x1)] =
1 1 1
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 0 0
1 0 1
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
[a__f(x1)] =
1 1 1
0 0 0
0 1 1
· x1 +
1 0 0
0 0 0
0 0 0
[h(x1)] =
1 0 0
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
0 0 0
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.