Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z077)

The rewrite relation of the following TRS is considered.

f(0(x1)) → s(0(x1)) (1)
d(0(x1)) → 0(x1) (2)
d(s(x1)) → s(s(d(x1))) (3)
f(s(x1)) → d(f(x1)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
0(f(x1)) → 0(s(x1)) (5)
0(d(x1)) → 0(x1) (6)
s(d(x1)) → d(s(s(x1))) (7)
s(f(x1)) → f(d(x1)) (8)

1.1 Rule Removal

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

1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
1 0 0
0 1 0
0 1 0
· x1 +
0 0 0
1 0 0
1 0 0
[0(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 0
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
0(d(x1)) → 0(x1) (6)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[d(x1)] =
1 0 0
0 1 1
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 1
0 0 1
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
s(f(x1)) → f(d(x1)) (8)

1.1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(d) = 0 weight(d) = 2
prec(s) = 1 weight(s) = 0
all of the following rules can be deleted.
s(d(x1)) → d(s(s(x1))) (7)

1.1.1.1.1.1 R is empty

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