MAYBE Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

Begin(c(c(c(x0)))) → Wait(Right1(x0))
Begin(c(c(x0))) → Wait(Right2(x0))
Begin(c(x0)) → Wait(Right3(x0))
Begin(b(x0)) → Wait(Right4(x0))
Begin(b(x0)) → Wait(Right5(x0))
Right1(c(End(x0))) → Left(b(b(b(b(End(x0))))))
Right2(c(c(End(x0)))) → Left(b(b(b(b(End(x0))))))
Right3(c(c(c(End(x0))))) → Left(b(b(b(b(End(x0))))))
Right4(b(End(x0))) → Left(End(x0))
Right5(b(End(x0))) → Left(c(b(c(End(x0)))))
Right1(c(x0)) → Ac(Right1(x0))
Right2(c(x0)) → Ac(Right2(x0))
Right3(c(x0)) → Ac(Right3(x0))
Right4(c(x0)) → Ac(Right4(x0))
Right5(c(x0)) → Ac(Right5(x0))
Right1(b(x0)) → Ab(Right1(x0))
Right2(b(x0)) → Ab(Right2(x0))
Right3(b(x0)) → Ab(Right3(x0))
Right4(b(x0)) → Ab(Right4(x0))
Right5(b(x0)) → Ab(Right5(x0))
Ac(Left(x0)) → Left(c(x0))
Ab(Left(x0)) → Left(b(x0))
Wait(Left(x0)) → Begin(x0)
c(c(c(c(x0)))) → b(b(b(b(x0))))
b(b(x0)) → x0
b(b(x0)) → c(b(c(x0)))

Proof

1 Termination Assumption

We assume termination of the following TRS
Begin(c(c(c(x0)))) → Wait(Right1(x0))
Begin(c(c(x0))) → Wait(Right2(x0))
Begin(c(x0)) → Wait(Right3(x0))
Begin(b(x0)) → Wait(Right4(x0))
Begin(b(x0)) → Wait(Right5(x0))
Right1(c(End(x0))) → Left(b(b(b(b(End(x0))))))
Right2(c(c(End(x0)))) → Left(b(b(b(b(End(x0))))))
Right3(c(c(c(End(x0))))) → Left(b(b(b(b(End(x0))))))
Right4(b(End(x0))) → Left(End(x0))
Right5(b(End(x0))) → Left(c(b(c(End(x0)))))
Right1(c(x0)) → Ac(Right1(x0))
Right2(c(x0)) → Ac(Right2(x0))
Right3(c(x0)) → Ac(Right3(x0))
Right4(c(x0)) → Ac(Right4(x0))
Right5(c(x0)) → Ac(Right5(x0))
Right1(b(x0)) → Ab(Right1(x0))
Right2(b(x0)) → Ab(Right2(x0))
Right3(b(x0)) → Ab(Right3(x0))
Right4(b(x0)) → Ab(Right4(x0))
Right5(b(x0)) → Ab(Right5(x0))
Ac(Left(x0)) → Left(c(x0))
Ab(Left(x0)) → Left(b(x0))
Wait(Left(x0)) → Begin(x0)
c(c(c(c(x0)))) → b(b(b(b(x0))))
b(b(x0)) → x0
b(b(x0)) → c(b(c(x0)))