Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/5-matchbox)

The rewrite relation of the following TRS is considered.

a(b(c(x1))) → a(a(b(x1))) (1)
a(b(c(x1))) → b(c(b(c(x1)))) (2)
a(b(c(x1))) → c(b(c(a(x1)))) (3)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(c(x1))) → a#(b(x1)) (4)
a#(b(c(x1))) → a#(a(b(x1))) (5)
a#(b(c(x1))) → a#(x1) (6)

1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[b(x1)] =
1 0 0
0 1 1
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 +
0 0 0
0 0 0
0 0 0
[c(x1)] =
0 0 1
0 1 0
1 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[a(x1)] =
0 0 0
1 1 0
1 1 0
· x1 +
1 0 0
0 0 0
0 0 0
together with the usable rules
a(b(c(x1))) → a(a(b(x1))) (1)
a(b(c(x1))) → b(c(b(c(x1)))) (2)
a(b(c(x1))) → c(b(c(a(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(b(c(x1))) → a#(b(x1)) (4)
a#(b(c(x1))) → a#(x1) (6)
could be deleted.

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over (2 x 2)-matrices with strict dimension 1 over the naturals
[b(x1)] =
0 0
1 0
· x1 +
0 0
0 0
[a#(x1)] =
0 2
0 0
· x1 +
0 0
0 0
[c(x1)] =
2 0
0 0
· x1 +
2 0
0 0
[a(x1)] =
0 1
0 2
· x1 +
0 0
0 0
together with the usable rules
a(b(c(x1))) → a(a(b(x1))) (1)
a(b(c(x1))) → b(c(b(c(x1)))) (2)
a(b(c(x1))) → c(b(c(a(x1)))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pair
a#(b(c(x1))) → a#(a(b(x1))) (5)
could be deleted.

1.1.1.1 P is empty

There are no pairs anymore.