Certification Problem

Input (TPDB TRS_Standard/Various_04/25)

The rewrite relation of the following TRS is considered.

g(a) → g(b) (1)
b → f(a,a) (2)
f(a,a) → g(d) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[d] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[g(x1)] =
1 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[f(x1, x2)] =
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x1 +
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[a] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[b] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
all of the following rules can be deleted.
g(a) → g(b) (1)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(d) = 0 weight(d) = 1
prec(f) = 2 weight(f) = 0
prec(b) = 4 weight(b) = 4
prec(g) = 7 weight(g) = 1
prec(a) = 6 weight(a) = 2
all of the following rules can be deleted.
b → f(a,a) (2)
f(a,a) → g(d) (3)

1.1.1 R is empty

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