Certification Problem

Input (TPDB SRS_Standard/Zantema_06/while2)

The rewrite relation of the following TRS is considered.

a(b(x1)) → d(x1) (1)
b(a(x1)) → a(b(x1)) (2)
d(c(x1)) → f(a(b(b(c(x1))))) (3)
d(f(x1)) → f(a(b(x1))) (4)
a(f(x1)) → a(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
b(a(x1)) → d(x1) (6)
a(b(x1)) → b(a(x1)) (7)
c(d(x1)) → c(b(b(a(f(x1))))) (8)
f(d(x1)) → b(a(f(x1))) (9)
f(a(x1)) → a(x1) (10)

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
f#(d(x1)) → f#(x1) (11)
f#(d(x1)) → b#(a(f(x1))) (12)
f#(d(x1)) → a#(f(x1)) (13)
c#(d(x1)) → f#(x1) (14)
c#(d(x1)) → c#(b(b(a(f(x1))))) (15)
c#(d(x1)) → b#(b(a(f(x1)))) (16)
c#(d(x1)) → b#(a(f(x1))) (17)
c#(d(x1)) → a#(f(x1)) (18)
a#(b(x1)) → b#(a(x1)) (19)
a#(b(x1)) → a#(x1) (20)

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
[f(x1)] = x1 +
0
[d(x1)] = x1 +
1
[c(x1)] = x1 +
0
[b(x1)] = x1 +
0
[a(x1)] = x1 +
1
[f#(x1)] = x1 +
2
[c#(x1)] = x1 +
2
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
2
together with the usable rules
b(a(x1)) → d(x1) (6)
a(b(x1)) → b(a(x1)) (7)
c(d(x1)) → c(b(b(a(f(x1))))) (8)
f(d(x1)) → b(a(f(x1))) (9)
f(a(x1)) → a(x1) (10)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
f#(d(x1)) → f#(x1) (11)
f#(d(x1)) → b#(a(f(x1))) (12)
f#(d(x1)) → a#(f(x1)) (13)
c#(d(x1)) → f#(x1) (14)
c#(d(x1)) → b#(b(a(f(x1)))) (16)
c#(d(x1)) → b#(a(f(x1))) (17)
c#(d(x1)) → a#(f(x1)) (18)
a#(b(x1)) → b#(a(x1)) (19)
and no rules could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.