Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.37a)

The rewrite relation of the following TRS is considered.

f(c(s(x),y)) → f(c(x,s(y))) (1)
g(c(x,s(y))) → g(c(s(x),y)) (2)
g(s(f(x))) → g(f(x)) (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
[c(x1, x2)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 1 0
0 0 0
0 0 1
· x2 +
0 0 0
1 0 0
0 0 0
[g(x1)] =
1 0 0
0 0 0
1 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 1 0
0 0 0
1 0 0
· x1 +
1 0 0
1 0 0
0 0 0
[f(x1)] =
1 0 0
1 0 0
1 0 0
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
g(s(f(x))) → g(f(x)) (3)

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[c(x1, x2)] =
1 0 1
0 0 0
0 1 1
· x1 +
1 1 1
0 0 1
0 0 1
· x2 +
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[s(x1)] =
1 0 1
0 1 1
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[f(x1)] =
1 0 1
0 0 0
1 0 1
· x1 +
0 0 0
0 0 0
0 0 0
all of the following rules can be deleted.
g(c(x,s(y))) → g(c(s(x),y)) (2)

1.1.1 Rule Removal

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

1.1.1.1 R is empty

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