Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
q0(a(x1)) → x(q1(x1))
q1(a(x1)) → a(q1(x1))
q1(y(x1)) → y(q1(x1))
a(q1(b(x1))) → q2(a(y(x1)))
a(q2(a(x1))) → q2(a(a(x1)))
a(q2(y(x1))) → q2(a(y(x1)))
y(q1(b(x1))) → q2(y(y(x1)))
y(q2(a(x1))) → q2(y(a(x1)))
y(q2(y(x1))) → q2(y(y(x1)))
q2(x(x1)) → x(q0(x1))
q0(y(x1)) → y(q3(x1))
q3(y(x1)) → y(q3(x1))
q3(bl(x1)) → bl(q4(x1))
Q is empty.
↳ QTRS
↳ DirectTerminationProof
Q restricted rewrite system:
The TRS R consists of the following rules:
q0(a(x1)) → x(q1(x1))
q1(a(x1)) → a(q1(x1))
q1(y(x1)) → y(q1(x1))
a(q1(b(x1))) → q2(a(y(x1)))
a(q2(a(x1))) → q2(a(a(x1)))
a(q2(y(x1))) → q2(a(y(x1)))
y(q1(b(x1))) → q2(y(y(x1)))
y(q2(a(x1))) → q2(y(a(x1)))
y(q2(y(x1))) → q2(y(y(x1)))
q2(x(x1)) → x(q0(x1))
q0(y(x1)) → y(q3(x1))
q3(y(x1)) → y(q3(x1))
q3(bl(x1)) → bl(q4(x1))
Q is empty.
We use [27] with the following order to prove termination.
Knuth-Bendix order [24] with precedence:
q11 > y1 > q21
q11 > a1 > q21
q31 > y1 > q21
and weight map:
q1_1=30
x_1=7
y_1=8
q0_1=27
bl_1=1
a_1=17
q4_1=9
q2_1=29
b_1=15
dummyConstant=1
q3_1=18