Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup14)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(a(b(b(b(b(a(a(x1)))))))) → b(b(a(a(c(c(a(a(x1)))))))) (4)
c(c(a(a(x1)))) → a(a(c(c(c(c(x1)))))) (5)
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1)))))) (3)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c#(c(c(c(c(c(x1)))))) → c#(c(b(b(x1)))) (6)
c#(c(c(c(c(c(x1)))))) → c#(b(b(x1))) (7)
c#(c(a(a(x1)))) → c#(x1) (8)
c#(c(a(a(x1)))) → c#(c(x1)) (9)
c#(c(a(a(x1)))) → c#(c(c(x1))) (10)
c#(c(a(a(x1)))) → c#(c(c(c(x1)))) (11)
c#(c(a(a(x1)))) → a#(c(c(c(c(x1))))) (12)
c#(c(a(a(x1)))) → a#(a(c(c(c(c(x1)))))) (13)
a#(a(b(b(b(b(a(a(x1)))))))) → c#(c(a(a(x1)))) (14)
a#(a(b(b(b(b(a(a(x1)))))))) → c#(a(a(x1))) (15)
a#(a(b(b(b(b(a(a(x1)))))))) → a#(c(c(a(a(x1))))) (16)
a#(a(b(b(b(b(a(a(x1)))))))) → a#(a(c(c(a(a(x1)))))) (17)

1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c(x1)] = x1 +
0
[b(x1)] = x1 +
0
[a(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[a#(x1)] = x1 +
0
together with the usable rules
a(a(b(b(b(b(a(a(x1)))))))) → b(b(a(a(c(c(a(a(x1)))))))) (4)
c(c(a(a(x1)))) → a(a(c(c(c(c(x1)))))) (5)
c(c(c(c(c(c(x1)))))) → b(b(c(c(b(b(x1)))))) (3)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c#(c(a(a(x1)))) → c#(x1) (8)
c#(c(a(a(x1)))) → c#(c(x1)) (9)
c#(c(a(a(x1)))) → c#(c(c(x1))) (10)
c#(c(a(a(x1)))) → c#(c(c(c(x1)))) (11)
c#(c(a(a(x1)))) → a#(c(c(c(c(x1))))) (12)
c#(c(a(a(x1)))) → a#(a(c(c(c(c(x1)))))) (13)
a#(a(b(b(b(b(a(a(x1)))))))) → c#(c(a(a(x1)))) (14)
a#(a(b(b(b(b(a(a(x1)))))))) → c#(a(a(x1))) (15)
a#(a(b(b(b(b(a(a(x1)))))))) → a#(c(c(a(a(x1))))) (16)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.