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

Proof

1 String Reversal

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

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals
[End(x1)] =
1 0 0
1 0 0
1 0 0
· x1 +
0 0 0
0 0 0
1 0 0
[Ab(x1)] =
1 0 0
0 0 0
0 1 1
· x1 +
0 0 0
0 0 0
0 0 0
[Begin(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[Wait(x1)] =
1 1 0
0 0 0
0 0 0
· x1 +
0 0 0
1 0 0
0 0 0
[b(x1)] =
1 0 0
0 1 0
0 0 0
· x1 +
0 0 0
0 0 0
0 0 0
[a(x1)] =
1 0 1
1 1 0
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[Aa(x1)] =
1 1 0
0 1 1
1 0 0
· x1 +
0 0 0
1 0 0
1 0 0
[Left(x1)] =
1 0 0
0 1 1
0 1 0
· x1 +
0 0 0
1 0 0
0 0 0
[Right1(x1)] =
1 0 0
0 1 1
0 1 0
· x1 +
1 0 0
1 0 0
0 0 0
the rules
a(Begin(x0)) → Right1(Wait(x0))
a(Right1(x0)) → Right1(Aa(x0))
b(Right1(x0)) → Right1(Ab(x0))
Left(Aa(x0)) → a(Left(x0))
Left(Ab(x0)) → b(Left(x0))
Left(Wait(x0)) → Begin(x0)
a(a(x0)) → a(b(a(x0)))
remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[Ab(x1)] = 0 · x1 + -∞
[Begin(x1)] = 6 · x1 + -∞
[Wait(x1)] = 8 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
[a(x1)] = 6 · x1 + -∞
[Aa(x1)] = 6 · x1 + -∞
[Left(x1)] = 4 · x1 + -∞
[Right1(x1)] = 4 · x1 + -∞
the rules
a(Begin(x0)) → Right1(Wait(x0))
a(Right1(x0)) → Right1(Aa(x0))
b(Right1(x0)) → Right1(Ab(x0))
Left(Aa(x0)) → a(Left(x0))
Left(Ab(x0)) → b(Left(x0))
a(a(x0)) → a(b(a(x0)))
remain.

1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers
[Ab(x1)] = 0 · x1 + -∞
[Begin(x1)] = 12 · x1 + -∞
[Wait(x1)] = 3 · x1 + -∞
[b(x1)] = 0 · x1 + -∞
[a(x1)] = 0 · x1 + -∞
[Aa(x1)] = 0 · x1 + -∞
[Left(x1)] = 0 · x1 + -∞
[Right1(x1)] = 8 · x1 + -∞
the rules
a(Right1(x0)) → Right1(Aa(x0))
b(Right1(x0)) → Right1(Ab(x0))
Left(Aa(x0)) → a(Left(x0))
Left(Ab(x0)) → b(Left(x0))
a(a(x0)) → a(b(a(x0)))
remain.

1.1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS
Right1(a(x0)) → Aa(Right1(x0))
Right1(b(x0)) → Ab(Right1(x0))
Aa(Left(x0)) → Left(a(x0))
Ab(Left(x0)) → Left(b(x0))
a(a(x0)) → a(b(a(x0)))

1.1.1.1.1.1 Bounds

The given TRS is match-bounded by 0. This is shown by the following automaton.