MAYBE Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

Begin(b(x0)) → Wait(Right1(x0))
Begin(c(x0)) → Wait(Right2(x0))
Right1(a(End(x0))) → Left(b(b(b(a(c(End(x0)))))))
Right2(c(End(x0))) → Left(a(End(x0)))
Right1(a(x0)) → Aa(Right1(x0))
Right2(a(x0)) → Aa(Right2(x0))
Right1(b(x0)) → Ab(Right1(x0))
Right2(b(x0)) → Ab(Right2(x0))
Right1(c(x0)) → Ac(Right1(x0))
Right2(c(x0)) → Ac(Right2(x0))
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Ac(Left(x0)) → Left(c(x0))
Wait(Left(x0)) → Begin(x0)
a(x0) → x0
a(b(x0)) → b(b(b(a(c(x0)))))
b(x0) → x0
c(c(x0)) → a(x0)

Proof

1 Termination Assumption

We assume termination of the following TRS
Begin(b(x0)) → Wait(Right1(x0))
Begin(c(x0)) → Wait(Right2(x0))
Right1(a(End(x0))) → Left(b(b(b(a(c(End(x0)))))))
Right2(c(End(x0))) → Left(a(End(x0)))
Right1(a(x0)) → Aa(Right1(x0))
Right2(a(x0)) → Aa(Right2(x0))
Right1(b(x0)) → Ab(Right1(x0))
Right2(b(x0)) → Ab(Right2(x0))
Right1(c(x0)) → Ac(Right1(x0))
Right2(c(x0)) → Ac(Right2(x0))
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Ac(Left(x0)) → Left(c(x0))
Wait(Left(x0)) → Begin(x0)
a(x0) → x0
a(b(x0)) → b(b(b(a(c(x0)))))
b(x0) → x0
c(c(x0)) → a(x0)