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 matchbox @ termCOMP 2023)

1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
e#(d(x1)) → e#(x1) (9)
e#(d(x1)) → c#(d(e(x1))) (10)
e#(d(x1)) → b#(c(d(e(x1)))) (11)
e#(d(x1)) → a#(b(c(d(e(x1))))) (12)
e#(c(x1)) → e#(x1) (13)
e#(c(x1)) → b#(a(a(e(x1)))) (14)
e#(c(x1)) → a#(e(x1)) (15)
e#(c(x1)) → a#(a(e(x1))) (16)
c#(d(d(x1))) → a#(x1) (17)
b#(b(x1)) → c#(x1) (18)
b#(b(x1)) → c#(c(x1)) (19)
b#(b(x1)) → c#(c(c(x1))) (20)
a#(x1) → c#(d(x1)) (21)
a#(a(x1)) → b#(x1) (22)
a#(a(x1)) → b#(b(x1)) (23)
a#(a(x1)) → b#(b(b(x1))) (24)

1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[e(x1)] = x1 +
0
[d(x1)] = x1 +
0
[c(x1)] = x1 +
0
[b(x1)] = x1 +
0
[a(x1)] = x1 +
0
[e#(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[b#(x1)] = x1 +
0
[a#(x1)] = x1 +
0
together with the usable rules
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)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
e#(d(x1)) → c#(d(e(x1))) (10)
e#(d(x1)) → b#(c(d(e(x1)))) (11)
e#(d(x1)) → a#(b(c(d(e(x1))))) (12)
e#(c(x1)) → b#(a(a(e(x1)))) (14)
e#(c(x1)) → a#(e(x1)) (15)
e#(c(x1)) → a#(a(e(x1))) (16)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.