Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x05)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(x1)) → b(b(x1)) (1)
b(c(c(x1))) → a(c(d(x1))) (5)
a(x1) → c(c(d(x1))) (6)
d(c(x1)) → c(b(x1)) (7)

1.1 Rule Removal

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

1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(c(c(x1))) → d#(x1) (8)
b#(c(c(x1))) → a#(c(d(x1))) (9)
a#(x1) → d#(x1) (10)
d#(c(x1)) → b#(x1) (11)

1.1.1.1 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)] =
0 1 0
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[b(x1)] =
0 1 0
0 1 0
0 1 0
· x1 +
0 0 0
0 0 0
0 0 0
[d(x1)] =
0 0 1
0 0 1
0 0 1
· x1 +
1 0 0
0 0 0
0 0 0
[d#(x1)] =
0 0 1
0 0 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
0 0 1
0 0 1
0 0 1
· x1 +
1 0 0
1 0 0
1 0 0
[a#(x1)] =
0 0 1
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
[c(x1)] =
0 0 1
1 0 0
0 1 0
· x1 +
1 0 0
0 0 0
0 0 0
together with the usable rules
b(c(c(x1))) → a(c(d(x1))) (5)
a(x1) → c(c(d(x1))) (6)
d(c(x1)) → c(b(x1)) (7)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(c(c(x1))) → d#(x1) (8)
a#(x1) → d#(x1) (10)
could be deleted.

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.