Certification Problem

Input (TPDB SRS_Standard/Secret_06_SRS/secr4)

The rewrite relation of the following TRS is considered.

a(a(x1)) → b(b(b(x1))) (1)
a(x1) → c(d(x1)) (2)
b(b(x1)) → c(c(c(x1))) (3)
c(c(x1)) → d(d(d(x1))) (4)
e(d(x1)) → a(b(c(d(e(x1))))) (5)
b(x1) → d(d(x1)) (6)
e(c(x1)) → b(a(a(e(x1)))) (7)
c(d(d(x1))) → a(x1) (8)

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.
e#(c(x1)) → b#(a(a(e(x1)))) (9)
e#(d(x1)) → a#(b(c(d(e(x1))))) (10)
a#(a(x1)) → b#(b(b(x1))) (11)
e#(d(x1)) → e#(x1) (12)
b#(b(x1)) → c#(c(x1)) (13)
b#(b(x1)) → c#(x1) (14)
a#(a(x1)) → b#(b(x1)) (15)
e#(d(x1)) → b#(c(d(e(x1)))) (16)
b#(b(x1)) → c#(c(c(x1))) (17)
e#(c(x1)) → e#(x1) (18)
c#(d(d(x1))) → a#(x1) (19)
e#(c(x1)) → a#(a(e(x1))) (20)
a#(x1) → c#(d(x1)) (21)
e#(d(x1)) → c#(d(e(x1))) (22)
a#(a(x1)) → b#(x1) (23)
e#(c(x1)) → a#(e(x1)) (24)

1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.