Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-210)

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

b(b(c(x1))) → b(a(a(x1))) (1)
a(c(c(x1))) → a(a(b(x1))) (2)

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), c(☐), a(☐)}

We obtain the transformed TRS
b(b(c(x1))) → b(a(a(x1))) (1)
a(c(c(x1))) → a(a(b(x1))) (2)
a(a(a(x1))) → a(b(a(x1))) (3)
a(b(a(x1))) → a(b(c(x1))) (6)
b(a(c(b(x1)))) → b(c(a(b(x1)))) (7)
c(a(c(b(x1)))) → c(c(a(b(x1)))) (8)
a(a(c(b(x1)))) → a(c(a(b(x1)))) (9)
b(a(c(c(x1)))) → b(c(c(b(x1)))) (10)
c(a(c(c(x1)))) → c(c(c(b(x1)))) (11)
a(a(c(c(x1)))) → a(c(c(b(x1)))) (12)

1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
bb(bc(cb(x1))) → ba(aa(ab(x1))) (13)
bb(bc(cc(x1))) → ba(aa(ac(x1))) (14)
bb(bc(ca(x1))) → ba(aa(aa(x1))) (15)
ac(cc(cb(x1))) → aa(ab(bb(x1))) (16)
ac(cc(cc(x1))) → aa(ab(bc(x1))) (17)
ac(cc(ca(x1))) → aa(ab(ba(x1))) (18)
aa(aa(ab(x1))) → ab(ba(ab(x1))) (19)
aa(aa(ac(x1))) → ab(ba(ac(x1))) (20)
aa(aa(aa(x1))) → ab(ba(aa(x1))) (21)
ab(ba(ab(x1))) → ab(bc(cb(x1))) (22)
ab(ba(ac(x1))) → ab(bc(cc(x1))) (23)
ab(ba(aa(x1))) → ab(bc(ca(x1))) (24)
ba(ac(cb(bb(x1)))) → bc(ca(ab(bb(x1)))) (25)
ba(ac(cb(bc(x1)))) → bc(ca(ab(bc(x1)))) (26)
ba(ac(cb(ba(x1)))) → bc(ca(ab(ba(x1)))) (27)
ca(ac(cb(bb(x1)))) → cc(ca(ab(bb(x1)))) (28)
ca(ac(cb(bc(x1)))) → cc(ca(ab(bc(x1)))) (29)
ca(ac(cb(ba(x1)))) → cc(ca(ab(ba(x1)))) (30)
aa(ac(cb(bb(x1)))) → ac(ca(ab(bb(x1)))) (31)
aa(ac(cb(bc(x1)))) → ac(ca(ab(bc(x1)))) (32)
aa(ac(cb(ba(x1)))) → ac(ca(ab(ba(x1)))) (33)
ba(ac(cc(cb(x1)))) → bc(cc(cb(bb(x1)))) (34)
ba(ac(cc(cc(x1)))) → bc(cc(cb(bc(x1)))) (35)
ba(ac(cc(ca(x1)))) → bc(cc(cb(ba(x1)))) (36)
ca(ac(cc(cb(x1)))) → cc(cc(cb(bb(x1)))) (37)
ca(ac(cc(cc(x1)))) → cc(cc(cb(bc(x1)))) (38)
ca(ac(cc(ca(x1)))) → cc(cc(cb(ba(x1)))) (39)
aa(ac(cc(cb(x1)))) → ac(cc(cb(bb(x1)))) (40)
aa(ac(cc(cc(x1)))) → ac(cc(cb(bc(x1)))) (41)
aa(ac(cc(ca(x1)))) → ac(cc(cb(ba(x1)))) (42)

1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(bb) = 2 weight(bb) = 8
prec(bc) = 0 weight(bc) = 10
prec(cb) = 6 weight(cb) = 5
prec(ba) = 1 weight(ba) = 14
prec(aa) = 7 weight(aa) = 8
prec(ab) = 8 weight(ab) = 1
prec(cc) = 3 weight(cc) = 9
prec(ac) = 5 weight(ac) = 5
prec(ca) = 4 weight(ca) = 12
all of the following rules can be deleted.
bb(bc(cb(x1))) → ba(aa(ab(x1))) (13)
bb(bc(cc(x1))) → ba(aa(ac(x1))) (14)
bb(bc(ca(x1))) → ba(aa(aa(x1))) (15)
ac(cc(cb(x1))) → aa(ab(bb(x1))) (16)
ac(cc(cc(x1))) → aa(ab(bc(x1))) (17)
ac(cc(ca(x1))) → aa(ab(ba(x1))) (18)
aa(aa(ab(x1))) → ab(ba(ab(x1))) (19)
aa(aa(ac(x1))) → ab(ba(ac(x1))) (20)
aa(aa(aa(x1))) → ab(ba(aa(x1))) (21)
ab(ba(ab(x1))) → ab(bc(cb(x1))) (22)
ab(ba(ac(x1))) → ab(bc(cc(x1))) (23)
ab(ba(aa(x1))) → ab(bc(ca(x1))) (24)
ba(ac(cb(bb(x1)))) → bc(ca(ab(bb(x1)))) (25)
ba(ac(cb(bc(x1)))) → bc(ca(ab(bc(x1)))) (26)
ba(ac(cb(ba(x1)))) → bc(ca(ab(ba(x1)))) (27)
ca(ac(cb(bb(x1)))) → cc(ca(ab(bb(x1)))) (28)
ca(ac(cb(bc(x1)))) → cc(ca(ab(bc(x1)))) (29)
ca(ac(cb(ba(x1)))) → cc(ca(ab(ba(x1)))) (30)
aa(ac(cb(bb(x1)))) → ac(ca(ab(bb(x1)))) (31)
aa(ac(cb(bc(x1)))) → ac(ca(ab(bc(x1)))) (32)
aa(ac(cb(ba(x1)))) → ac(ca(ab(ba(x1)))) (33)
ba(ac(cc(cb(x1)))) → bc(cc(cb(bb(x1)))) (34)
ba(ac(cc(cc(x1)))) → bc(cc(cb(bc(x1)))) (35)
ba(ac(cc(ca(x1)))) → bc(cc(cb(ba(x1)))) (36)
ca(ac(cc(cb(x1)))) → cc(cc(cb(bb(x1)))) (37)
ca(ac(cc(cc(x1)))) → cc(cc(cb(bc(x1)))) (38)
ca(ac(cc(ca(x1)))) → cc(cc(cb(ba(x1)))) (39)
aa(ac(cc(cb(x1)))) → ac(cc(cb(bb(x1)))) (40)
aa(ac(cc(cc(x1)))) → ac(cc(cb(bc(x1)))) (41)
aa(ac(cc(ca(x1)))) → ac(cc(cb(ba(x1)))) (42)

1.1.1.1 R is empty

There are no rules in the TRS. Hence, it is terminating.