Certification Problem

Input (TPDB TRS_Standard/Zantema_05/z26)

The rewrite relation of the following TRS is considered.

a(a(f(x,y))) → f(a(b(a(b(a(x))))),a(b(a(b(a(y)))))) (1)
f(a(x),a(y)) → a(f(x,y)) (2)
f(b(x),b(y)) → b(f(x,y)) (3)

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
[a(x1)] =
1 0 1
0 0 0
0 0 1
· x1 +
0 0 0
0 0 0
1 0 0
[f(x1, x2)] =
1 0 0
0 0 0
0 0 1
· x1 +
1 0 0
0 0 0
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
a(a(f(x,y))) → f(a(b(a(b(a(x))))),a(b(a(b(a(y)))))) (1)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(b) = 2 weight(b) = 2
prec(a) = 0 weight(a) = 2
prec(f) = 1 weight(f) = 0
all of the following rules can be deleted.
f(a(x),a(y)) → a(f(x,y)) (2)
f(b(x),b(y)) → b(f(x,y)) (3)

1.1.1 R is empty

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