Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-66)

The rewrite relation of the following TRS is considered.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by matchbox @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c(x1)] = x1 +
0
[b(x1)] = x1 +
0
[a(x1)] = x1 +
1
all of the following rules can be deleted.
b(b(a(x1))) → b(b(b(x1))) (4)
a(c(b(x1))) → c(b(c(x1))) (5)
a(c(b(x1))) → b(b(c(x1))) (6)

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(c(a(b(x1)))) → c(a(b(c(x1)))) (7)
c(a(b(a(x1)))) → c(c(a(a(x1)))) (8)
c(c(c(c(x1)))) → c(b(c(c(x1)))) (9)
b(c(a(b(x1)))) → b(a(b(c(x1)))) (10)
b(a(b(a(x1)))) → b(c(a(a(x1)))) (11)
b(c(c(c(x1)))) → b(b(c(c(x1)))) (12)
a(c(a(b(x1)))) → a(a(b(c(x1)))) (13)
a(a(b(a(x1)))) → a(c(a(a(x1)))) (14)
a(c(c(c(x1)))) → a(b(c(c(x1)))) (15)

1.1.1 Semantic Labeling

The following interpretations form a model of the rules.

As carrier we take the set {0,1,2}. Symbols are labeled by the interpretation of their arguments using the interpretations (modulo 3):

[c(x1)] = 3x1 + 0
[b(x1)] = 3x1 + 1
[a(x1)] = 3x1 + 2

We obtain the labeled TRS
c0(c2(a1(b0(x1)))) → c2(a1(b0(c0(x1)))) (16)
c0(c2(a1(b2(x1)))) → c2(a1(b0(c2(x1)))) (17)
c0(c2(a1(b1(x1)))) → c2(a1(b0(c1(x1)))) (18)
a0(c2(a1(b0(x1)))) → a2(a1(b0(c0(x1)))) (19)
a0(c2(a1(b2(x1)))) → a2(a1(b0(c2(x1)))) (20)
a0(c2(a1(b1(x1)))) → a2(a1(b0(c1(x1)))) (21)
b0(c2(a1(b0(x1)))) → b2(a1(b0(c0(x1)))) (22)
b0(c2(a1(b2(x1)))) → b2(a1(b0(c2(x1)))) (23)
b0(c2(a1(b1(x1)))) → b2(a1(b0(c1(x1)))) (24)
c2(a1(b2(a0(x1)))) → c0(c2(a2(a0(x1)))) (25)
c2(a1(b2(a2(x1)))) → c0(c2(a2(a2(x1)))) (26)
c2(a1(b2(a1(x1)))) → c0(c2(a2(a1(x1)))) (27)
a2(a1(b2(a0(x1)))) → a0(c2(a2(a0(x1)))) (28)
a2(a1(b2(a2(x1)))) → a0(c2(a2(a2(x1)))) (29)
a2(a1(b2(a1(x1)))) → a0(c2(a2(a1(x1)))) (30)
b2(a1(b2(a0(x1)))) → b0(c2(a2(a0(x1)))) (31)
b2(a1(b2(a2(x1)))) → b0(c2(a2(a2(x1)))) (32)
b2(a1(b2(a1(x1)))) → b0(c2(a2(a1(x1)))) (33)
c0(c0(c0(c0(x1)))) → c1(b0(c0(c0(x1)))) (34)
c0(c0(c0(c2(x1)))) → c1(b0(c0(c2(x1)))) (35)
c0(c0(c0(c1(x1)))) → c1(b0(c0(c1(x1)))) (36)
a0(c0(c0(c0(x1)))) → a1(b0(c0(c0(x1)))) (37)
a0(c0(c0(c2(x1)))) → a1(b0(c0(c2(x1)))) (38)
a0(c0(c0(c1(x1)))) → a1(b0(c0(c1(x1)))) (39)
b0(c0(c0(c0(x1)))) → b1(b0(c0(c0(x1)))) (40)
b0(c0(c0(c2(x1)))) → b1(b0(c0(c2(x1)))) (41)
b0(c0(c0(c1(x1)))) → b1(b0(c0(c1(x1)))) (42)

1.1.1.1 Rule Removal

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c0(x1)] = x1 +
3/2
[c1(x1)] = x1 +
0
[c2(x1)] = x1 +
3/2
[b0(x1)] = x1 +
1
[b1(x1)] = x1 +
1/2
[b2(x1)] = x1 +
1
[a0(x1)] = x1 +
1
[a1(x1)] = x1 +
3/2
[a2(x1)] = x1 +
0
all of the following rules can be deleted.
c0(c2(a1(b1(x1)))) → c2(a1(b0(c1(x1)))) (18)
a0(c2(a1(b0(x1)))) → a2(a1(b0(c0(x1)))) (19)
a0(c2(a1(b2(x1)))) → a2(a1(b0(c2(x1)))) (20)
a0(c2(a1(b1(x1)))) → a2(a1(b0(c1(x1)))) (21)
b0(c2(a1(b1(x1)))) → b2(a1(b0(c1(x1)))) (24)
c2(a1(b2(a0(x1)))) → c0(c2(a2(a0(x1)))) (25)
c2(a1(b2(a2(x1)))) → c0(c2(a2(a2(x1)))) (26)
c2(a1(b2(a1(x1)))) → c0(c2(a2(a1(x1)))) (27)
b2(a1(b2(a0(x1)))) → b0(c2(a2(a0(x1)))) (31)
b2(a1(b2(a2(x1)))) → b0(c2(a2(a2(x1)))) (32)
b2(a1(b2(a1(x1)))) → b0(c2(a2(a1(x1)))) (33)
c0(c0(c0(c0(x1)))) → c1(b0(c0(c0(x1)))) (34)
c0(c0(c0(c2(x1)))) → c1(b0(c0(c2(x1)))) (35)
c0(c0(c0(c1(x1)))) → c1(b0(c0(c1(x1)))) (36)
b0(c0(c0(c0(x1)))) → b1(b0(c0(c0(x1)))) (40)
b0(c0(c0(c2(x1)))) → b1(b0(c0(c2(x1)))) (41)
b0(c0(c0(c1(x1)))) → b1(b0(c0(c1(x1)))) (42)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
c0#(c2(a1(b0(x1)))) → c0#(x1) (43)
c0#(c2(a1(b0(x1)))) → b0#(c0(x1)) (44)
c0#(c2(a1(b2(x1)))) → b0#(c2(x1)) (45)
b0#(c2(a1(b0(x1)))) → c0#(x1) (46)
b0#(c2(a1(b0(x1)))) → b0#(c0(x1)) (47)
b0#(c2(a1(b2(x1)))) → b0#(c2(x1)) (48)
a0#(c0(c0(c0(x1)))) → b0#(c0(c0(x1))) (49)
a0#(c0(c0(c1(x1)))) → b0#(c0(c1(x1))) (50)
a0#(c0(c0(c2(x1)))) → b0#(c0(c2(x1))) (51)
a2#(a1(b2(a0(x1)))) → a0#(c2(a2(a0(x1)))) (52)
a2#(a1(b2(a0(x1)))) → a2#(a0(x1)) (53)
a2#(a1(b2(a1(x1)))) → a0#(c2(a2(a1(x1)))) (54)
a2#(a1(b2(a1(x1)))) → a2#(a1(x1)) (55)
a2#(a1(b2(a2(x1)))) → a0#(c2(a2(a2(x1)))) (56)
a2#(a1(b2(a2(x1)))) → a2#(a2(x1)) (57)

1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules

Using the matrix interpretations of dimension 1 with strict dimension 1 over the rationals with delta = 1
[c0(x1)] = x1 +
1
[c1(x1)] = x1 +
0
[c2(x1)] = x1 +
1
[b0(x1)] = x1 +
1
[b2(x1)] = x1 +
1
[a0(x1)] = x1 +
1
[a1(x1)] = x1 +
1
[a2(x1)] = x1 +
0
[c0#(x1)] = x1 +
1
[b0#(x1)] = x1 +
0
[a0#(x1)] = x1 +
1
[a2#(x1)] = x1 +
1
together with the usable rules
c0(c2(a1(b0(x1)))) → c2(a1(b0(c0(x1)))) (16)
c0(c2(a1(b2(x1)))) → c2(a1(b0(c2(x1)))) (17)
b0(c2(a1(b0(x1)))) → b2(a1(b0(c0(x1)))) (22)
b0(c2(a1(b2(x1)))) → b2(a1(b0(c2(x1)))) (23)
a2(a1(b2(a0(x1)))) → a0(c2(a2(a0(x1)))) (28)
a2(a1(b2(a2(x1)))) → a0(c2(a2(a2(x1)))) (29)
a2(a1(b2(a1(x1)))) → a0(c2(a2(a1(x1)))) (30)
a0(c0(c0(c0(x1)))) → a1(b0(c0(c0(x1)))) (37)
a0(c0(c0(c2(x1)))) → a1(b0(c0(c2(x1)))) (38)
a0(c0(c0(c1(x1)))) → a1(b0(c0(c1(x1)))) (39)
(w.r.t. the implicit argument filter of the reduction pair), the pairs
c0#(c2(a1(b0(x1)))) → c0#(x1) (43)
c0#(c2(a1(b0(x1)))) → b0#(c0(x1)) (44)
c0#(c2(a1(b2(x1)))) → b0#(c2(x1)) (45)
b0#(c2(a1(b0(x1)))) → c0#(x1) (46)
b0#(c2(a1(b0(x1)))) → b0#(c0(x1)) (47)
b0#(c2(a1(b2(x1)))) → b0#(c2(x1)) (48)
a0#(c0(c0(c0(x1)))) → b0#(c0(c0(x1))) (49)
a0#(c0(c0(c1(x1)))) → b0#(c0(c1(x1))) (50)
a0#(c0(c0(c2(x1)))) → b0#(c0(c2(x1))) (51)
a2#(a1(b2(a0(x1)))) → a0#(c2(a2(a0(x1)))) (52)
a2#(a1(b2(a0(x1)))) → a2#(a0(x1)) (53)
a2#(a1(b2(a1(x1)))) → a0#(c2(a2(a1(x1)))) (54)
a2#(a1(b2(a1(x1)))) → a2#(a1(x1)) (55)
a2#(a1(b2(a2(x1)))) → a0#(c2(a2(a2(x1)))) (56)
a2#(a1(b2(a2(x1)))) → a2#(a2(x1)) (57)
and no rules could be deleted.

1.1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 0 components.