Certification Problem

Input (TPDB SRS_Standard/Wenzel_16/baabababba-abababbaababab.srs)

The rewrite relation of the following TRS is considered.

b(a(a(b(a(b(a(b(b(a(x1)))))))))) → a(b(a(b(a(b(b(a(a(b(a(b(a(b(x1)))))))))))))) (1)

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(b(b(a(b(a(b(a(a(b(x1)))))))))) → b(a(b(a(b(a(a(b(b(a(b(a(b(a(x1)))))))))))))) (2)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(x1) (3)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(x1))) (4)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(b(a(x1))))) (5)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(b(a(b(a(b(a(x1)))))))) (6)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(a(b(b(a(b(a(b(a(x1))))))))) (7)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(a(b(b(a(b(a(b(a(x1))))))))))) (8)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(b(a(a(b(b(a(b(a(b(a(x1))))))))))))) (9)

1.1.1 Reduction Pair Processor with Usable Rules

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 1 · x1 + -∞
[a(x1)] = -1 · x1 + 1
[a#(x1)] = 0 · x1 + -16
together with the usable rule
a(b(b(a(b(a(b(a(a(b(x1)))))))))) → b(a(b(a(b(a(a(b(b(a(b(a(b(a(x1)))))))))))))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(x1) (3)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(x1))) (4)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(b(a(x1))))) (5)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(a(b(b(a(b(a(b(a(x1))))))))) (7)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(a(b(b(a(b(a(b(a(x1))))))))))) (8)
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(a(b(a(a(b(b(a(b(a(b(a(x1))))))))))))) (9)
could be deleted.

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 0 1
1 0 0
0 0 1
· x1 +
0 0 0
1 0 0
0 0 0
[a(x1)] =
0 1 0
1 1 1
1 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a#(x1)] =
0 1 0
0 0 0
0 0 0
· x1 +
1 0 0
0 0 0
0 0 0
together with the usable rule
a(b(b(a(b(a(b(a(a(b(x1)))))))))) → b(a(b(a(b(a(a(b(b(a(b(a(b(a(x1)))))))))))))) (2)
(w.r.t. the implicit argument filter of the reduction pair), the pair
a#(b(b(a(b(a(b(a(a(b(x1)))))))))) → a#(b(b(a(b(a(b(a(x1)))))))) (6)
could be deleted.

1.1.1.1.1 P is empty

There are no pairs anymore.