Certification Problem

Input (TPDB SRS_Standard/Secret_07_SRS/x07)

The rewrite relation of the following TRS is considered.

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

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#(e(e(x1))) → d#(e(x1)) (10)
d#(x1) → e#(x1) (11)
d#(x1) → b#(e(x1)) (12)
d#(c(x1)) → d#(a(x1)) (13)
d#(c(x1)) → c#(d(a(x1))) (14)
d#(c(x1)) → a#(x1) (15)
c#(d(a(x1))) → c#(x1) (16)
b#(e(b(x1))) → e#(d(x1)) (17)
b#(e(b(x1))) → d#(x1) (18)
b#(d(x1)) → e#(b(x1)) (19)
b#(d(x1)) → b#(x1) (20)
b#(b(b(x1))) → e#(x1) (21)
b#(b(b(x1))) → e#(e(x1)) (22)
a#(x1) → e#(b(x1)) (23)
a#(x1) → b#(x1) (24)
a#(a(b(x1))) → d#(x1) (25)
a#(a(b(x1))) → c#(d(x1)) (26)

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
[d#(x1)] = x1 +
1
[c#(x1)] = x1 +
0
[b#(x1)] = x1 +
1
[a#(x1)] = x1 +
1
together with the usable rules
a(a(b(x1))) → c(d(x1)) (1)
b(e(b(x1))) → e(d(x1)) (2)
b(d(x1)) → e(b(x1)) (3)
b(b(b(x1))) → e(e(x1)) (4)
e(e(e(x1))) → d(e(x1)) (5)
d(x1) → b(e(x1)) (6)
c(d(a(x1))) → c(x1) (7)
d(c(x1)) → c(d(a(x1))) (8)
a(x1) → e(b(x1)) (9)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
d#(c(x1)) → c#(d(a(x1))) (14)
a#(a(b(x1))) → c#(d(x1)) (26)
and no rules could be deleted.

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.