Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-227)

The relative rewrite relation R/S is considered where R is the following TRS

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

and S is the following TRS.

c(b(a(x1))) → b(a(c(x1))) (6)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Rule Removal

Using the matrix interpretations of dimension 2 with strict dimension 1 over the integers
[c(x1)] =
0
2
+
2 0
0 2
· x1
[b(x1)] =
2
0
+
2 2
0 0
· x1
[a(x1)] =
2
0
+
2 0
0 0
· x1
all of the following rules can be deleted.
c(b(b(x1))) → a(a(c(x1))) (2)
c(b(a(x1))) → b(a(c(x1))) (6)

1.1 Closure Under Flat Contexts

Using the flat contexts

{c(☐), b(☐), a(☐)}

We obtain the transformed TRS
c(b(c(x1))) → c(a(b(x1))) (1)
b(b(b(x1))) → b(a(b(x1))) (5)
c(a(a(b(x1)))) → c(b(b(c(x1)))) (7)
b(a(a(b(x1)))) → b(b(b(c(x1)))) (8)
a(a(a(b(x1)))) → a(b(b(c(x1)))) (9)
c(b(b(c(x1)))) → c(a(b(c(x1)))) (10)
b(b(b(c(x1)))) → b(a(b(c(x1)))) (11)
a(b(b(c(x1)))) → a(a(b(c(x1)))) (12)

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
cb(bc(cc(x1))) → ca(ab(bc(x1))) (13)
cb(bc(cb(x1))) → ca(ab(bb(x1))) (14)
cb(bc(ca(x1))) → ca(ab(ba(x1))) (15)
bb(bb(bc(x1))) → ba(ab(bc(x1))) (16)
bb(bb(bb(x1))) → ba(ab(bb(x1))) (17)
bb(bb(ba(x1))) → ba(ab(ba(x1))) (18)
ca(aa(ab(bc(x1)))) → cb(bb(bc(cc(x1)))) (19)
ca(aa(ab(bb(x1)))) → cb(bb(bc(cb(x1)))) (20)
ca(aa(ab(ba(x1)))) → cb(bb(bc(ca(x1)))) (21)
ba(aa(ab(bc(x1)))) → bb(bb(bc(cc(x1)))) (22)
ba(aa(ab(bb(x1)))) → bb(bb(bc(cb(x1)))) (23)
ba(aa(ab(ba(x1)))) → bb(bb(bc(ca(x1)))) (24)
aa(aa(ab(bc(x1)))) → ab(bb(bc(cc(x1)))) (25)
aa(aa(ab(bb(x1)))) → ab(bb(bc(cb(x1)))) (26)
aa(aa(ab(ba(x1)))) → ab(bb(bc(ca(x1)))) (27)
cb(bb(bc(cc(x1)))) → ca(ab(bc(cc(x1)))) (28)
cb(bb(bc(cb(x1)))) → ca(ab(bc(cb(x1)))) (29)
cb(bb(bc(ca(x1)))) → ca(ab(bc(ca(x1)))) (30)
bb(bb(bc(cc(x1)))) → ba(ab(bc(cc(x1)))) (31)
bb(bb(bc(cb(x1)))) → ba(ab(bc(cb(x1)))) (32)
bb(bb(bc(ca(x1)))) → ba(ab(bc(ca(x1)))) (33)
ab(bb(bc(cc(x1)))) → aa(ab(bc(cc(x1)))) (34)
ab(bb(bc(cb(x1)))) → aa(ab(bc(cb(x1)))) (35)
ab(bb(bc(ca(x1)))) → aa(ab(bc(ca(x1)))) (36)

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals
[cb(x1)] = 1 · x1 + 1
[bc(x1)] = 1 · x1
[cc(x1)] = 1 · x1
[ca(x1)] = 1 · x1 + 1
[ab(x1)] = 1 · x1
[bb(x1)] = 1 · x1 + 1
[ba(x1)] = 1 · x1 + 1
[aa(x1)] = 1 · x1 + 1
all of the following rules can be deleted.
bb(bb(bc(x1))) → ba(ab(bc(x1))) (16)
bb(bb(bb(x1))) → ba(ab(bb(x1))) (17)
bb(bb(ba(x1))) → ba(ab(ba(x1))) (18)
aa(aa(ab(bc(x1)))) → ab(bb(bc(cc(x1)))) (25)
aa(aa(ab(bb(x1)))) → ab(bb(bc(cb(x1)))) (26)
aa(aa(ab(ba(x1)))) → ab(bb(bc(ca(x1)))) (27)
cb(bb(bc(cc(x1)))) → ca(ab(bc(cc(x1)))) (28)
cb(bb(bc(cb(x1)))) → ca(ab(bc(cb(x1)))) (29)
cb(bb(bc(ca(x1)))) → ca(ab(bc(ca(x1)))) (30)
bb(bb(bc(cc(x1)))) → ba(ab(bc(cc(x1)))) (31)
bb(bb(bc(cb(x1)))) → ba(ab(bc(cb(x1)))) (32)
bb(bb(bc(ca(x1)))) → ba(ab(bc(ca(x1)))) (33)

1.1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
cb#(bc(cc(x1))) → ca#(ab(bc(x1))) (37)
cb#(bc(cc(x1))) → ab#(bc(x1)) (38)
cb#(bc(cb(x1))) → ca#(ab(bb(x1))) (39)
cb#(bc(cb(x1))) → ab#(bb(x1)) (40)
cb#(bc(ca(x1))) → ca#(ab(ba(x1))) (41)
cb#(bc(ca(x1))) → ab#(ba(x1)) (42)
cb#(bc(ca(x1))) → ba#(x1) (43)
ca#(aa(ab(bc(x1)))) → cb#(bb(bc(cc(x1)))) (44)
ca#(aa(ab(bb(x1)))) → cb#(bb(bc(cb(x1)))) (45)
ca#(aa(ab(bb(x1)))) → cb#(x1) (46)
ca#(aa(ab(ba(x1)))) → cb#(bb(bc(ca(x1)))) (47)
ca#(aa(ab(ba(x1)))) → ca#(x1) (48)
ba#(aa(ab(bb(x1)))) → cb#(x1) (49)
ba#(aa(ab(ba(x1)))) → ca#(x1) (50)
ab#(bb(bc(cc(x1)))) → ab#(bc(cc(x1))) (51)
ab#(bb(bc(cb(x1)))) → ab#(bc(cb(x1))) (52)
ab#(bb(bc(ca(x1)))) → ab#(bc(ca(x1))) (53)

1.1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.