Certification Problem

Input (TPDB TRS_Standard/Transformed_CSR_04/Ex9_Luc06_GM)

The rewrite relation of the following TRS is considered.

a__f(a,X,X) → a__f(X,a__b,b) (1)
a__b → a (2)
mark(f(X1,X2,X3)) → a__f(X1,mark(X2),X3) (3)
mark(b) → a__b (4)
mark(a) → a (5)
a__f(X1,X2,X3) → f(X1,X2,X3) (6)
a__b → b (7)

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
[f(x1, x2, x3)] =
1 1 1
0 0 0
0 0 0
· x1 +
1 0 1
0 0 0
0 0 0
· x2 +
1 1 0
0 0 0
0 0 0
· x3 +
0 0 0
1 0 0
1 0 0
[a__f(x1, x2, x3)] =
1 1 1
0 0 0
0 0 0
· x1 +
1 0 1
0 0 0
0 0 0
· x2 +
1 1 0
0 0 0
0 0 0
· x3 +
0 0 0
1 0 0
1 0 0
[b] =
0 0 0
0 0 0
0 0 0
[mark(x1)] =
1 0 1
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
1 0 0
[a] =
0 0 0
1 0 0
1 0 0
[a__b] =
1 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
a__b → a (2)
mark(a) → a (5)
a__b → b (7)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(mark) = 7 weight(mark) = 2
prec(f) = 0 weight(f) = 0
prec(b) = 6 weight(b) = 1
prec(a__b) = 5 weight(a__b) = 3
prec(a__f) = 4 weight(a__f) = 0
prec(a) = 2 weight(a) = 6
all of the following rules can be deleted.
a__f(a,X,X) → a__f(X,a__b,b) (1)
mark(f(X1,X2,X3)) → a__f(X1,mark(X2),X3) (3)
mark(b) → a__b (4)
a__f(X1,X2,X3) → f(X1,X2,X3) (6)

1.1.1 R is empty

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