Certification Problem

Input (TPDB SRS_Standard/Secret_05_SRS/matchbox2)

The rewrite relation of the following TRS is considered.

t(o(x1)) m(a(x1)) (1)
t(e(x1)) n(s(x1)) (2)
a(l(x1)) a(t(x1)) (3)
o(m(a(x1))) t(e(n(x1))) (4)
s(a(x1)) l(a(t(o(m(a(t(e(x1)))))))) (5)
n(s(x1)) a(l(a(t(x1)))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by NaTT @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
t#(e(x1)) s#(x1) (7)
s#(a(x1)) a#(t(o(m(a(t(e(x1))))))) (8)
t#(o(x1)) a#(x1) (9)
s#(a(x1)) a#(t(e(x1))) (10)
a#(l(x1)) a#(t(x1)) (11)
a#(l(x1)) t#(x1) (12)
o#(m(a(x1))) t#(e(n(x1))) (13)
s#(a(x1)) t#(o(m(a(t(e(x1)))))) (14)
s#(a(x1)) t#(e(x1)) (15)
n#(s(x1)) t#(x1) (16)
n#(s(x1)) a#(l(a(t(x1)))) (17)
t#(e(x1)) n#(s(x1)) (18)
s#(a(x1)) o#(m(a(t(e(x1))))) (19)
o#(m(a(x1))) n#(x1) (20)
n#(s(x1)) a#(t(x1)) (21)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.