Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z119)

The rewrite relation of the following TRS is considered.

a(b(x1)) b(d(x1)) (1)
a(c(x1)) d(d(d(x1))) (2)
b(d(x1)) a(c(b(x1))) (3)
c(f(x1)) d(d(c(x1))) (4)
d(d(x1)) f(x1) (5)
f(f(x1)) a(x1) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
a#(b(x1)) b#(d(x1)) (7)
a#(b(x1)) d#(x1) (8)
a#(c(x1)) d#(d(d(x1))) (9)
a#(c(x1)) d#(d(x1)) (10)
a#(c(x1)) d#(x1) (11)
b#(d(x1)) a#(c(b(x1))) (12)
b#(d(x1)) c#(b(x1)) (13)
b#(d(x1)) b#(x1) (14)
c#(f(x1)) d#(d(c(x1))) (15)
c#(f(x1)) d#(c(x1)) (16)
c#(f(x1)) c#(x1) (17)
d#(d(x1)) f#(x1) (18)
f#(f(x1)) a#(x1) (19)

1.1 Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a#(x1)] = 1 · x1
[b(x1)] = 1 + 1 · x1
[b#(x1)] = 1 + 1 · x1
[d(x1)] = 1 · x1
[d#(x1)] = 1 · x1
[c(x1)] = 1 · x1
[c#(x1)] = 1 · x1
[f(x1)] = 1 · x1
[f#(x1)] = 1 · x1
[a(x1)] = 1 · x1
the pair
a#(b(x1)) d#(x1) (8)
could be deleted.

1.1.1 Monotonic Reduction Pair Processor

Using the linear polynomial interpretation over the naturals
[a(x1)] = 3 + 1 · x1
[b(x1)] = 3 · x1
[d(x1)] = 1 + 1 · x1
[c(x1)] = 1 · x1
[f(x1)] = 2 + 1 · x1
[a#(x1)] = 3 + 1 · x1
[b#(x1)] = 3 · x1
[d#(x1)] = 1 · x1
[c#(x1)] = 2 + 1 · x1
[f#(x1)] = 1 + 1 · x1
the pairs
a#(c(x1)) d#(d(d(x1))) (9)
a#(c(x1)) d#(d(x1)) (10)
a#(c(x1)) d#(x1) (11)
b#(d(x1)) c#(b(x1)) (13)
b#(d(x1)) b#(x1) (14)
c#(f(x1)) d#(d(c(x1))) (15)
c#(f(x1)) d#(c(x1)) (16)
c#(f(x1)) c#(x1) (17)
and the rule
f(f(x1)) a(x1) (6)
could be deleted.

1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.