Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/torpa1)

The rewrite relation of the following TRS is considered.

a(d(x1)) → d(b(c(b(d(x1))))) (1)
a(x1) → b(b(f(b(b(x1))))) (2)
b(d(b(x1))) → a(d(x1)) (3)
d(f(x1)) → b(d(x1)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Rule Removal

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[f(x1)] =
1 0
0 1
· x1 +
4 0
1 0
[a(x1)] =
1 0
0 1
· x1 +
4 0
5 0
[c(x1)] =
1 0
0 0
· x1 +
0 0
0 0
[d(x1)] =
1 4
0 4
· x1 +
0 0
3 0
[b(x1)] =
1 0
0 1
· x1 +
0 0
1 0
all of the following rules can be deleted.
d(f(x1)) → b(d(x1)) (4)

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
[a(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
1 0 0
1 0 0
1 0 0
[c(x1)] =
1 0 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
1 1 1
0 0 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
1 0 0
0 0 0
all of the following rules can be deleted.
a(x1) → b(b(f(b(b(x1))))) (2)

1.1.1 Rule Removal

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

1.1.1.1 Dependency Pair Transformation

The set of initial dependency pairs is empty, and hence the TRS is terminating.