Certification Problem

Input (TPDB SRS_Relative/Waldmann_19/random-35)

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

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

and S is the following TRS.

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

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by AProVE @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
b(c(b(x1))) → c(c(a(x1))) (6)
b(b(c(x1))) → b(c(c(x1))) (7)
b(c(a(x1))) → c(a(b(x1))) (8)
a(c(a(x1))) → a(a(b(x1))) (9)
a(a(c(x1))) → b(b(c(x1))) (10)

1.1 Closure Under Flat Contexts

Using the flat contexts

{b(☐), c(☐), a(☐)}

We obtain the transformed TRS
b(b(c(x1))) → b(c(c(x1))) (7)
b(b(c(b(x1)))) → b(c(c(a(x1)))) (11)
c(b(c(b(x1)))) → c(c(c(a(x1)))) (12)
a(b(c(b(x1)))) → a(c(c(a(x1)))) (13)
a(c(a(x1))) → a(a(b(x1))) (9)
b(b(c(a(x1)))) → b(c(a(b(x1)))) (14)
c(b(c(a(x1)))) → c(c(a(b(x1)))) (15)
a(b(c(a(x1)))) → a(c(a(b(x1)))) (16)
b(a(a(c(x1)))) → b(b(b(c(x1)))) (17)
c(a(a(c(x1)))) → c(b(b(c(x1)))) (18)
a(a(a(c(x1)))) → a(b(b(c(x1)))) (19)

1.1.1 Semantic Labeling

Root-labeling is applied.

We obtain the labeled TRS
bb(bc(cb(x1))) → bc(cc(cb(x1))) (20)
bb(bc(cc(x1))) → bc(cc(cc(x1))) (21)
bb(bc(ca(x1))) → bc(cc(ca(x1))) (22)
bb(bc(cb(bb(x1)))) → bc(cc(ca(ab(x1)))) (23)
bb(bc(cb(bc(x1)))) → bc(cc(ca(ac(x1)))) (24)
bb(bc(cb(ba(x1)))) → bc(cc(ca(aa(x1)))) (25)
cb(bc(cb(bb(x1)))) → cc(cc(ca(ab(x1)))) (26)
cb(bc(cb(bc(x1)))) → cc(cc(ca(ac(x1)))) (27)
cb(bc(cb(ba(x1)))) → cc(cc(ca(aa(x1)))) (28)
ab(bc(cb(bb(x1)))) → ac(cc(ca(ab(x1)))) (29)
ab(bc(cb(bc(x1)))) → ac(cc(ca(ac(x1)))) (30)
ab(bc(cb(ba(x1)))) → ac(cc(ca(aa(x1)))) (31)
ac(ca(ab(x1))) → aa(ab(bb(x1))) (32)
ac(ca(ac(x1))) → aa(ab(bc(x1))) (33)
ac(ca(aa(x1))) → aa(ab(ba(x1))) (34)
bb(bc(ca(ab(x1)))) → bc(ca(ab(bb(x1)))) (35)
bb(bc(ca(ac(x1)))) → bc(ca(ab(bc(x1)))) (36)
bb(bc(ca(aa(x1)))) → bc(ca(ab(ba(x1)))) (37)
cb(bc(ca(ab(x1)))) → cc(ca(ab(bb(x1)))) (38)
cb(bc(ca(ac(x1)))) → cc(ca(ab(bc(x1)))) (39)
cb(bc(ca(aa(x1)))) → cc(ca(ab(ba(x1)))) (40)
ab(bc(ca(ab(x1)))) → ac(ca(ab(bb(x1)))) (41)
ab(bc(ca(ac(x1)))) → ac(ca(ab(bc(x1)))) (42)
ab(bc(ca(aa(x1)))) → ac(ca(ab(ba(x1)))) (43)
ba(aa(ac(cb(x1)))) → bb(bb(bc(cb(x1)))) (44)
ba(aa(ac(cc(x1)))) → bb(bb(bc(cc(x1)))) (45)
ba(aa(ac(ca(x1)))) → bb(bb(bc(ca(x1)))) (46)
ca(aa(ac(cb(x1)))) → cb(bb(bc(cb(x1)))) (47)
ca(aa(ac(cc(x1)))) → cb(bb(bc(cc(x1)))) (48)
ca(aa(ac(ca(x1)))) → cb(bb(bc(ca(x1)))) (49)
aa(aa(ac(cb(x1)))) → ab(bb(bc(cb(x1)))) (50)
aa(aa(ac(cc(x1)))) → ab(bb(bc(cc(x1)))) (51)
aa(aa(ac(ca(x1)))) → ab(bb(bc(ca(x1)))) (52)

1.1.1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(bb) = 1 weight(bb) = 4
prec(bc) = 0 weight(bc) = 5
prec(cb) = 5 weight(cb) = 4
prec(cc) = 2 weight(cc) = 1
prec(ca) = 6 weight(ca) = 7
prec(ab) = 4 weight(ab) = 2
prec(ac) = 3 weight(ac) = 3
prec(ba) = 8 weight(ba) = 6
prec(aa) = 7 weight(aa) = 5
all of the following rules can be deleted.
bb(bc(cb(x1))) → bc(cc(cb(x1))) (20)
bb(bc(cc(x1))) → bc(cc(cc(x1))) (21)
bb(bc(ca(x1))) → bc(cc(ca(x1))) (22)
bb(bc(cb(bb(x1)))) → bc(cc(ca(ab(x1)))) (23)
bb(bc(cb(bc(x1)))) → bc(cc(ca(ac(x1)))) (24)
bb(bc(cb(ba(x1)))) → bc(cc(ca(aa(x1)))) (25)
cb(bc(cb(bb(x1)))) → cc(cc(ca(ab(x1)))) (26)
cb(bc(cb(bc(x1)))) → cc(cc(ca(ac(x1)))) (27)
cb(bc(cb(ba(x1)))) → cc(cc(ca(aa(x1)))) (28)
ab(bc(cb(bb(x1)))) → ac(cc(ca(ab(x1)))) (29)
ab(bc(cb(bc(x1)))) → ac(cc(ca(ac(x1)))) (30)
ab(bc(cb(ba(x1)))) → ac(cc(ca(aa(x1)))) (31)
ac(ca(ab(x1))) → aa(ab(bb(x1))) (32)
ac(ca(ac(x1))) → aa(ab(bc(x1))) (33)
ac(ca(aa(x1))) → aa(ab(ba(x1))) (34)
bb(bc(ca(ab(x1)))) → bc(ca(ab(bb(x1)))) (35)
bb(bc(ca(ac(x1)))) → bc(ca(ab(bc(x1)))) (36)
bb(bc(ca(aa(x1)))) → bc(ca(ab(ba(x1)))) (37)
cb(bc(ca(ab(x1)))) → cc(ca(ab(bb(x1)))) (38)
cb(bc(ca(ac(x1)))) → cc(ca(ab(bc(x1)))) (39)
cb(bc(ca(aa(x1)))) → cc(ca(ab(ba(x1)))) (40)
ab(bc(ca(ab(x1)))) → ac(ca(ab(bb(x1)))) (41)
ab(bc(ca(ac(x1)))) → ac(ca(ab(bc(x1)))) (42)
ab(bc(ca(aa(x1)))) → ac(ca(ab(ba(x1)))) (43)
ba(aa(ac(cb(x1)))) → bb(bb(bc(cb(x1)))) (44)
ba(aa(ac(cc(x1)))) → bb(bb(bc(cc(x1)))) (45)
ba(aa(ac(ca(x1)))) → bb(bb(bc(ca(x1)))) (46)
ca(aa(ac(cb(x1)))) → cb(bb(bc(cb(x1)))) (47)
ca(aa(ac(cc(x1)))) → cb(bb(bc(cc(x1)))) (48)
ca(aa(ac(ca(x1)))) → cb(bb(bc(ca(x1)))) (49)
aa(aa(ac(cb(x1)))) → ab(bb(bc(cb(x1)))) (50)
aa(aa(ac(cc(x1)))) → ab(bb(bc(cc(x1)))) (51)
aa(aa(ac(ca(x1)))) → ab(bb(bc(ca(x1)))) (52)

1.1.1.1.1 R is empty

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