Certification Problem

Input (TPDB SRS_Standard/Trafo_06/dup16)

The rewrite relation of the following TRS is considered.

a(a(a(a(x1)))) → b(b(b(b(b(b(x1)))))) (1)
b(b(b(b(x1)))) → c(c(c(c(c(c(x1)))))) (2)
c(c(c(c(x1)))) → d(d(d(d(d(d(x1)))))) (3)
b(b(x1)) → d(d(d(d(x1)))) (4)
c(c(d(d(d(d(x1)))))) → a(a(x1)) (5)

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.
b#(b(b(b(x1)))) → c#(c(c(c(c(x1))))) (6)
b#(b(b(b(x1)))) → c#(x1) (7)
a#(a(a(a(x1)))) → b#(b(x1)) (8)
a#(a(a(a(x1)))) → b#(b(b(b(b(b(x1)))))) (9)
a#(a(a(a(x1)))) → b#(b(b(b(x1)))) (10)
a#(a(a(a(x1)))) → b#(b(b(x1))) (11)
a#(a(a(a(x1)))) → b#(x1) (12)
c#(c(d(d(d(d(x1)))))) → a#(a(x1)) (13)
a#(a(a(a(x1)))) → b#(b(b(b(b(x1))))) (14)
b#(b(b(b(x1)))) → c#(c(x1)) (15)
b#(b(b(b(x1)))) → c#(c(c(c(x1)))) (16)
b#(b(b(b(x1)))) → c#(c(c(c(c(c(x1)))))) (17)
c#(c(d(d(d(d(x1)))))) → a#(x1) (18)
b#(b(b(b(x1)))) → c#(c(c(x1))) (19)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.