YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 3 · x1 + -∞
[Right4(x1)] = 4 · x1 + -∞
[Begin(x1)] = 0 · x1 + -∞
[Wait(x1)] = 0 · x1 + -∞
[Right2(x1)] = 3 · x1 + -∞
[End(x1)] = 4 · x1 + -∞
[a(x1)] = 5 · x1 + -∞
[Right3(x1)] = 6 · x1 + -∞
[Ac(x1)] = 2 · x1 + -∞
[Aa(x1)] = 5 · x1 + -∞
[Left(x1)] = 0 · x1 + -∞
[Right5(x1)] = 2 · x1 + -∞
[Ab(x1)] = 3 · x1 + -∞
[c(x1)] = 2 · x1 + -∞
[Right1(x1)] = 4 · x1 + -∞
the rules
Begin(b(x0)) → Wait(Right2(x0))
Begin(c(c(c(x0)))) → Wait(Right3(x0))
Begin(c(c(x0))) → Wait(Right4(x0))
Begin(c(x0)) → Wait(Right5(x0))
Right1(a(End(x0))) → Left(b(b(b(End(x0)))))
Right2(b(End(x0))) → Left(c(c(c(End(x0)))))
Right3(c(End(x0))) → Left(a(b(End(x0))))
Right4(c(c(End(x0)))) → Left(a(b(End(x0))))
Right5(c(c(c(End(x0))))) → Left(a(b(End(x0))))
Right1(a(x0)) → Aa(Right1(x0))
Right2(a(x0)) → Aa(Right2(x0))
Right3(a(x0)) → Aa(Right3(x0))
Right4(a(x0)) → Aa(Right4(x0))
Right5(a(x0)) → Aa(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))
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))
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Ac(Left(x0)) → Left(c(x0))
Wait(Left(x0)) → Begin(x0)
b(b(x0)) → c(c(c(x0)))
c(c(c(c(x0)))) → a(b(x0))
remain.

1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 0 · x1 + -∞
[Right4(x1)] = 0 · x1 + -∞
[Begin(x1)] = 0 · x1 + -∞
[Wait(x1)] = 0 · x1 + -∞
[Right2(x1)] = 0 · x1 + -∞
[End(x1)] = 0 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
[Right3(x1)] = 0 · x1 + -∞
[Ac(x1)] = 0 · x1 + -∞
[Aa(x1)] = 0 · x1 + -∞
[Left(x1)] = 0 · x1 + -∞
[Right5(x1)] = 0 · x1 + -∞
[Ab(x1)] = 0 · x1 + -∞
[c(x1)] = 0 · x1 + -∞
[Right1(x1)] = 6 · x1 + -∞
the rules
Begin(b(x0)) → Wait(Right2(x0))
Begin(c(c(c(x0)))) → Wait(Right3(x0))
Begin(c(c(x0))) → Wait(Right4(x0))
Begin(c(x0)) → Wait(Right5(x0))
Right2(b(End(x0))) → Left(c(c(c(End(x0)))))
Right3(c(End(x0))) → Left(a(b(End(x0))))
Right4(c(c(End(x0)))) → Left(a(b(End(x0))))
Right5(c(c(c(End(x0))))) → Left(a(b(End(x0))))
Right1(a(x0)) → Aa(Right1(x0))
Right2(a(x0)) → Aa(Right2(x0))
Right3(a(x0)) → Aa(Right3(x0))
Right4(a(x0)) → Aa(Right4(x0))
Right5(a(x0)) → Aa(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))
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))
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Ac(Left(x0)) → Left(c(x0))
Wait(Left(x0)) → Begin(x0)
b(b(x0)) → c(c(c(x0)))
c(c(c(c(x0)))) → a(b(x0))
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[b(x1)] = 5 · x1 + -∞
[Right4(x1)] = 1 · x1 + -∞
[Begin(x1)] = 0 · x1 + -∞
[Wait(x1)] = 0 · x1 + -∞
[Right2(x1)] = 1 · x1 + -∞
[End(x1)] = 7 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
[Right3(x1)] = 3 · x1 + -∞
[Ac(x1)] = 2 · x1 + -∞
[Aa(x1)] = 0 · x1 + -∞
[Left(x1)] = 0 · x1 + -∞
[Right5(x1)] = 2 · x1 + -∞
[Ab(x1)] = 5 · x1 + -∞
[c(x1)] = 2 · x1 + -∞
[Right1(x1)] = 0 · x1 + -∞
the rules
Begin(c(x0)) → Wait(Right5(x0))
Right2(b(End(x0))) → Left(c(c(c(End(x0)))))
Right3(c(End(x0))) → Left(a(b(End(x0))))
Right4(c(c(End(x0)))) → Left(a(b(End(x0))))
Right1(a(x0)) → Aa(Right1(x0))
Right2(a(x0)) → Aa(Right2(x0))
Right3(a(x0)) → Aa(Right3(x0))
Right4(a(x0)) → Aa(Right4(x0))
Right5(a(x0)) → Aa(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))
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))
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
Ac(Left(x0)) → Left(c(x0))
Wait(Left(x0)) → Begin(x0)
remain.

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
c(Begin(x0)) → Right5(Wait(x0))
End(b(Right2(x0))) → End(c(c(c(Left(x0)))))
End(c(Right3(x0))) → End(b(a(Left(x0))))
End(c(c(Right4(x0)))) → End(b(a(Left(x0))))
a(Right1(x0)) → Right1(Aa(x0))
a(Right2(x0)) → Right2(Aa(x0))
a(Right3(x0)) → Right3(Aa(x0))
a(Right4(x0)) → Right4(Aa(x0))
a(Right5(x0)) → Right5(Aa(x0))
b(Right1(x0)) → Right1(Ab(x0))
b(Right2(x0)) → Right2(Ab(x0))
b(Right3(x0)) → Right3(Ab(x0))
b(Right4(x0)) → Right4(Ab(x0))
b(Right5(x0)) → Right5(Ab(x0))
c(Right1(x0)) → Right1(Ac(x0))
c(Right2(x0)) → Right2(Ac(x0))
c(Right3(x0)) → Right3(Ac(x0))
c(Right4(x0)) → Right4(Ac(x0))
c(Right5(x0)) → Right5(Ac(x0))
Left(Aa(x0)) → a(Left(x0))
Left(Ab(x0)) → b(Left(x0))
Left(Ac(x0)) → c(Left(x0))
Left(Wait(x0)) → Begin(x0)

1.1.1.1.1 Bounds

The given TRS is match-bounded by 1. This is shown by the following automaton.