Certification Problem

Input (TPDB SRS_Standard/Zantema_06/12)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(l(x1)) a#(x1) (5)
a#(l(x1)) l#(a(x1)) (6)
a#(c(x1)) a#(x1) (7)
a#(c(x1)) c#(a(x1)) (8)
c#(a(r(x1))) a#(x1) (9)
l#(r(a(a(x1)))) c#(r(x1)) (10)
l#(r(a(a(x1)))) c#(c(r(x1))) (11)
l#(r(a(a(x1)))) c#(c(c(r(x1)))) (12)
l#(r(a(a(x1)))) l#(c(c(c(r(x1))))) (13)
l#(r(a(a(x1)))) a#(l(c(c(c(r(x1)))))) (14)
l#(r(a(a(x1)))) a#(a(l(c(c(c(r(x1))))))) (15)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.