Certification Problem

Input (TPDB TRS_Relative/INVY_15/invNSS03)

The relative rewrite relation R/S is considered where R is the following TRS

f(x,0) → s(x) (1)
g(x) → h(x,gen) (2)
h(0,x) → f(x,x) (3)
a → b (4)

and S is the following TRS.

gen → s(gen) (5)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over the naturals
[gen] = 4
[f(x1, x2)] = 1 · x1 + 2 · x2 + 17
[g(x1)] = 18 · x1 + 18
[h(x1, x2)] = 7 · x1 + 4 · x2 + 2
[0] = 9
[b] = 0
[a] = 1
[s(x1)] = 1 · x1 + 0
all of the following rules can be deleted.
f(x,0) → s(x) (1)
h(0,x) → f(x,x) (3)
a → b (4)

1.1 Rule Removal

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals
[gen] =
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
[g(x1)] =
1 1 1 1 1
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
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
1 0 0 0 0
[h(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 1 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
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.
g(x) → h(x,gen) (2)

1.1.1 R is empty

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