Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z072)

The rewrite relation of the following TRS is considered.

a(l(x1)) l(a(x1)) (1)
r(a(a(x1))) a(a(r(x1))) (2)
b(l(x1)) b(a(r(x1))) (3)
r(b(x1)) l(b(x1)) (4)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(l(x1)) b#(a(r(x1))) (5)
b#(l(x1)) a#(r(x1)) (6)
b#(l(x1)) r#(x1) (7)
a#(l(x1)) a#(x1) (8)
r#(a(a(x1))) a#(a(r(x1))) (9)
r#(a(a(x1))) a#(r(x1)) (10)
r#(a(a(x1))) r#(x1) (11)

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
[l(x1)] = x1 +
1
[b(x1)] = x1 +
0
[a(x1)] = x1 +
0
[r(x1)] = x1 +
1
[b#(x1)] = x1 +
2
[a#(x1)] = x1 +
0
[r#(x1)] = x1 +
2
together with the usable rules
a(l(x1)) l(a(x1)) (1)
r(a(a(x1))) a(a(r(x1))) (2)
b(l(x1)) b(a(r(x1))) (3)
r(b(x1)) l(b(x1)) (4)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
b#(l(x1)) a#(r(x1)) (6)
b#(l(x1)) r#(x1) (7)
a#(l(x1)) a#(x1) (8)
r#(a(a(x1))) a#(a(r(x1))) (9)
r#(a(a(x1))) a#(r(x1)) (10)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.