Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/jambox5)

The rewrite relation of the following TRS is considered.

b(a(x1)) → a(a(d(x1))) (1)
a(c(x1)) → b(b(x1)) (2)
d(a(b(x1))) → b(d(d(c(x1)))) (3)
d(x1) → a(x1) (4)
b(a(c(a(x1)))) → x1 (5)

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(b(x1)) → d(a(a(x1))) (6)
c(a(x1)) → b(b(x1)) (7)
b(a(d(x1))) → c(d(d(b(x1)))) (8)
d(x1) → a(x1) (4)
a(c(a(b(x1)))) → x1 (9)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
d#(x1) → a#(x1) (10)
c#(a(x1)) → b#(x1) (11)
c#(a(x1)) → b#(b(x1)) (12)
b#(a(d(x1))) → d#(d(b(x1))) (13)
b#(a(d(x1))) → d#(b(x1)) (14)
b#(a(d(x1))) → c#(d(d(b(x1)))) (15)
b#(a(d(x1))) → b#(x1) (16)
a#(b(x1)) → d#(a(a(x1))) (17)
a#(b(x1)) → a#(x1) (18)
a#(b(x1)) → a#(a(x1)) (19)

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
[d(x1)] = x1 +
0
[c(x1)] = x1 +
0
[b(x1)] = x1 +
0
[a(x1)] = x1 +
0
[d#(x1)] = x1 +
0
[c#(x1)] = x1 +
1
[b#(x1)] = x1 +
1
[a#(x1)] = x1 +
0
together with the usable rules
a(b(x1)) → d(a(a(x1))) (6)
c(a(x1)) → b(b(x1)) (7)
b(a(d(x1))) → c(d(d(b(x1)))) (8)
d(x1) → a(x1) (4)
a(c(a(b(x1)))) → x1 (9)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(a(d(x1))) → d#(d(b(x1))) (13)
b#(a(d(x1))) → d#(b(x1)) (14)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.