Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex23_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(f(a)) → c(f(g(f(a)))) (1)
mark(f(X)) → a__f(mark(X)) (2)
mark(a) → a (3)
mark(c(X)) → c(X) (4)
mark(g(X)) → g(mark(X)) (5)
a__f(X) → f(X) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[f(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
1 0 0
[g(x1)] =
1 0 1
0 1 1
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[mark(x1)] =
1 1 1
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[a] =
1 0 0
0 0 0
0 0 0
[a__f(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
1 0 0
0 0 0
1 0 0
all of the following rules can be deleted.
mark(a) → a (3)
mark(c(X)) → c(X) (4)
a__f(X) → f(X) (6)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[f(x1)] =
1 1 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[g(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[mark(x1)] =
1 1 1
0 0 0
0 0 1
· x1 +
0 0 0
1 0 0
1 0 0
[a] =
0 0 0
1 0 0
0 0 0
[a__f(x1)] =
1 0 0
0 1 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
mark(g(X)) → g(mark(X)) (5)

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
1 0 0
[f(x1)] =
1 1 1
1 1 1
1 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[g(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[mark(x1)] =
1 1 1
1 0 1
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
[a] =
0 0 0
0 0 0
0 0 0
[a__f(x1)] =
1 1 1
1 0 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
mark(f(X)) → a__f(mark(X)) (2)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(c) = 0 weight(c) = 2
prec(g) = 5 weight(g) = 2
prec(a__f) = 1 weight(a__f) = 5
prec(f) = 2 weight(f) = 1
prec(a) = 4 weight(a) = 2
all of the following rules can be deleted.
a__f(f(a)) → c(f(g(f(a)))) (1)

1.1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.