Certification Problem

Input (TPDB TRS_Standard/Rubio_04/test4)

The rewrite relation of the following TRS is considered.

f(a,a) → f(a,b) (1)
f(a,b) → f(s(a),c) (2)
f(s(X),c) → f(X,c) (3)
f(c,c) → f(a,a) (4)

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
[c] =
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[f(x1, x2)] =
1 0 1 1 0
0 0 0 0 0
0 0 0 1 0
0 0 0 0 0
0 0 1 1 0
· x1 +
1 0 0 0 1
0 0 0 0 0
0 1 0 0 0
0 0 0 0 0
0 0 0 0 1
· x2 +
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[s(x1)] =
1 0 0 0 0
0 0 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 0
· x1 +
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
0 0 0 0 0
[a] =
0 0 0 0 0
1 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
1 0 0 0 0
all of the following rules can be deleted.
f(s(X),c) → f(X,c) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[c] =
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
[f(x1, x2)] =
1 1 0 0 0
0 0 0 1 1
0 0 0 0 0
0 0 0 0 0
0 0 0 0 0
· 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
· 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
[s(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
· x1 +
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
[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.
f(a,a) → f(a,b) (1)

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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