Certification Problem

Input (TPDB TRS_Standard/Strategy_removed_AG01/#4.21)

The rewrite relation of the following TRS is considered.

f(1) → f(g(1)) (1)
f(f(x)) → f(x) (2)
g(0) → g(f(0)) (3)
g(g(x)) → g(x) (4)

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
[f(x1)] =
1 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
1 0 0
[1] =
0 0 0
0 0 0
0 0 0
[g(x1)] =
1 1 1
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
all of the following rules can be deleted.
f(f(x)) → f(x) (2)
g(g(x)) → g(x) (4)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0 1 0 0
0 0 0 0 1
0 0 0 0 0
0 0 0 0 0
0 1 1 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
[0] =
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
[1] =
0 0 0 0 0
0 0 0 0 0
1 0 0 0 0
0 0 0 0 0
1 0 0 0 0
[g(x1)] =
1 0 0 1 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
1 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(1) → f(g(1)) (1)
g(0) → g(f(0)) (3)

1.1.1 R is empty

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