Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x06)

The rewrite relation of the following TRS is considered.

a(b(x1)) → c(b(x1)) (1)
c(c(x1)) → d(b(x1)) (2)
d(x1) → c(e(x1)) (3)
b(b(x1)) → f(x1) (4)
c(b(x1)) → g(x1) (5)
e(x1) → f(x1) (6)
e(x1) → b(b(x1)) (7)
f(g(x1)) → a(c(x1)) (8)
g(f(x1)) → e(x1) (9)
a(x1) → b(c(x1)) (10)

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.
g#(f(x1)) → e#(x1) (11)
f#(g(x1)) → c#(x1) (12)
f#(g(x1)) → a#(c(x1)) (13)
e#(x1) → f#(x1) (14)
e#(x1) → b#(x1) (15)
e#(x1) → b#(b(x1)) (16)
d#(x1) → e#(x1) (17)
d#(x1) → c#(e(x1)) (18)
c#(c(x1)) → d#(b(x1)) (19)
c#(c(x1)) → b#(x1) (20)
c#(b(x1)) → g#(x1) (21)
b#(b(x1)) → f#(x1) (22)
a#(x1) → c#(x1) (23)
a#(x1) → b#(c(x1)) (24)
a#(b(x1)) → c#(b(x1)) (25)

1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.