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.