Certification Problem

Input (TPDB SRS_Standard/Zantema_04/z124)

The rewrite relation of the following TRS is considered.

q0(a(x1)) x(q1(x1)) (1)
q1(a(x1)) a(q1(x1)) (2)
q1(y(x1)) y(q1(x1)) (3)
a(q1(b(x1))) q2(a(y(x1))) (4)
a(q2(a(x1))) q2(a(a(x1))) (5)
a(q2(y(x1))) q2(a(y(x1))) (6)
y(q1(b(x1))) q2(y(y(x1))) (7)
y(q2(a(x1))) q2(y(a(x1))) (8)
y(q2(y(x1))) q2(y(y(x1))) (9)
q2(x(x1)) x(q0(x1)) (10)
q0(y(x1)) y(q3(x1)) (11)
q3(y(x1)) y(q3(x1)) (12)
q3(bl(x1)) bl(q4(x1)) (13)

Property / Task

Prove or disprove termination.

Answer / Result

Yes.

Proof (by ttt2 @ termCOMP 2023)

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
a(q0(x1)) q1(x(x1)) (14)
a(q1(x1)) q1(a(x1)) (15)
y(q1(x1)) q1(y(x1)) (16)
b(q1(a(x1))) y(a(q2(x1))) (17)
a(q2(a(x1))) a(a(q2(x1))) (18)
y(q2(a(x1))) y(a(q2(x1))) (19)
b(q1(y(x1))) y(y(q2(x1))) (20)
a(q2(y(x1))) a(y(q2(x1))) (21)
y(q2(y(x1))) y(y(q2(x1))) (22)
x(q2(x1)) q0(x(x1)) (23)
y(q0(x1)) q3(y(x1)) (24)
y(q3(x1)) q3(y(x1)) (25)
bl(q3(x1)) q4(bl(x1)) (26)

1.1 Rule Removal

Using the Knuth Bendix order with w0 = 1 and the following precedence and weight functions
prec(q4) = 2 weight(q4) = 2
prec(bl) = 11 weight(bl) = 2
prec(q3) = 4 weight(q3) = 2
prec(q2) = 7 weight(q2) = 5
prec(b) = 14 weight(b) = 2
prec(y) = 5 weight(y) = 2
prec(x) = 10 weight(x) = 1
prec(q1) = 0 weight(q1) = 5
prec(q0) = 15 weight(q0) = 3
prec(a) = 1 weight(a) = 3
all of the following rules can be deleted.
a(q0(x1)) q1(x(x1)) (14)
a(q1(x1)) q1(a(x1)) (15)
y(q1(x1)) q1(y(x1)) (16)
b(q1(a(x1))) y(a(q2(x1))) (17)
a(q2(a(x1))) a(a(q2(x1))) (18)
y(q2(a(x1))) y(a(q2(x1))) (19)
b(q1(y(x1))) y(y(q2(x1))) (20)
a(q2(y(x1))) a(y(q2(x1))) (21)
y(q2(y(x1))) y(y(q2(x1))) (22)
x(q2(x1)) q0(x(x1)) (23)
y(q0(x1)) q3(y(x1)) (24)
y(q3(x1)) q3(y(x1)) (25)
bl(q3(x1)) q4(bl(x1)) (26)

1.1.1 R is empty

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