YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

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

1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

[Wait(x₁)]

1	0	0
0	1	0
0	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[End(x₁)]

1	1	0
0	0	0
1	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[a(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Right1(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Right2(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Ab(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[b(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
1	0	0
0	0	0

[Right4(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Aa(x₁)]

1	0	0
0	0	0
0	1	1

· x₁ +

0	0	0
0	0	0
0	0	0

[Left(x₁)]

1	0	0
0	1	1
0	1	1

· x₁ +

0	0	0
1	0	0
0	0	0

[Right3(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Begin(x₁)]

1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

the rules

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

remain.

1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers

[Wait(x₁)]	=	11 · x₁ + -∞
[End(x₁)]	=	9 · x₁ + -∞
[a(x₁)]	=	0 · x₁ + -∞
[Right1(x₁)]	=	0 · x₁ + -∞
[Right2(x₁)]	=	10 · x₁ + -∞
[Ab(x₁)]	=	4 · x₁ + -∞
[b(x₁)]	=	4 · x₁ + -∞
[Right4(x₁)]	=	0 · x₁ + -∞
[Aa(x₁)]	=	0 · x₁ + -∞
[Left(x₁)]	=	2 · x₁ + -∞
[Right3(x₁)]	=	10 · x₁ + -∞
[Begin(x₁)]	=	13 · x₁ + -∞

the rules

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

remain.

1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

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

1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[Wait(x₁)]	=	1 · x₁ + 2
[End(x₁)]	=	12 · x₁ + 2
[a(x₁)]	=	1 · x₁ + 0
[Right1(x₁)]	=	4 · x₁ + 12
[Right2(x₁)]	=	1 · x₁ + 10
[Ab(x₁)]	=	1 · x₁ + 4
[b(x₁)]	=	1 · x₁ + 4
[Right4(x₁)]	=	4 · x₁ + 0
[Aa(x₁)]	=	1 · x₁ + 0
[Left(x₁)]	=	1 · x₁ + 2
[Right3(x₁)]	=	1 · x₁ + 10
[Begin(x₁)]	=	1 · x₁ + 4

the rules

Begin(a(b(b(x0))))	→	Wait(Right2(x0))
Begin(b(b(x0)))	→	Wait(Right3(x0))
Right2(b(a(End(x0))))	→	Left(a(a(b(b(b(a(a(End(x0)))))))))
Right3(b(a(a(End(x0)))))	→	Left(a(a(b(b(b(a(a(End(x0)))))))))
Right2(b(x0))	→	Ab(Right2(x0))
Right3(b(x0))	→	Ab(Right3(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))
Ab(Left(x0))	→	Left(b(x0))
Aa(Left(x0))	→	Left(a(x0))
Wait(Left(x0))	→	Begin(x0)
b(a(a(b(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

remain.

1.1.1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

b(b(a(Begin(x0))))	→	Right2(Wait(x0))
b(b(Begin(x0)))	→	Right3(Wait(x0))
End(a(b(Right2(x0))))	→	End(a(a(b(b(b(a(a(Left(x0)))))))))
End(a(a(b(Right3(x0)))))	→	End(a(a(b(b(b(a(a(Left(x0)))))))))
b(Right2(x0))	→	Right2(Ab(x0))
b(Right3(x0))	→	Right3(Ab(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))
Left(Ab(x0))	→	b(Left(x0))
Left(Aa(x0))	→	a(Left(x0))
Left(Wait(x0))	→	Begin(x0)
b(b(a(a(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

[Wait(x₁)]

1	1	1
1	0	1
1	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[End(x₁)]

1	0	1
0	0	0
1	0	1

· x₁ +

0	0	0
0	0	0
0	0	0

[a(x₁)]

1	0	0
0	0	0
0	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Right1(x₁)]

1	0	1
0	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Right2(x₁)]

1	0	0
0	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Ab(x₁)]

1	0	0
0	0	1
1	0	1

· x₁ +

0	0	0
0	0	0
1	0	0

[b(x₁)]

1	0	0
1	1	0
0	0	0

· x₁ +

0	0	0
1	0	0
0	0	0

[Right4(x₁)]

1	1	1
0	0	0
0	0	0

· x₁ +

1	0	0
0	0	0
0	0	0

[Aa(x₁)]

1	0	0
0	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Left(x₁)]

1	0	0
0	0	1
0	1	0

· x₁ +

0	0	0
1	0	0
1	0	0

[Right3(x₁)]

1	0	0
0	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Begin(x₁)]

1	1	1
0	1	0
1	0	0

· x₁ +

0	0	0
1	0	0
0	0	0

the rules

b(b(a(Begin(x0))))	→	Right2(Wait(x0))
b(b(Begin(x0)))	→	Right3(Wait(x0))
End(a(a(b(Right3(x0)))))	→	End(a(a(b(b(b(a(a(Left(x0)))))))))
b(Right2(x0))	→	Right2(Ab(x0))
b(Right3(x0))	→	Right3(Ab(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))
Left(Ab(x0))	→	b(Left(x0))
Left(Aa(x0))	→	a(Left(x0))
Left(Wait(x0))	→	Begin(x0)
b(b(a(a(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

remain.

1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers

[Wait(x₁)]	=	0 · x₁ + -∞
[End(x₁)]	=	0 · x₁ + -∞
[a(x₁)]	=	0 · x₁ + -∞
[Right1(x₁)]	=	8 · x₁ + -∞
[Right2(x₁)]	=	0 · x₁ + -∞
[Ab(x₁)]	=	0 · x₁ + -∞
[b(x₁)]	=	0 · x₁ + -∞
[Right4(x₁)]	=	9 · x₁ + -∞
[Aa(x₁)]	=	0 · x₁ + -∞
[Left(x₁)]	=	1 · x₁ + -∞
[Right3(x₁)]	=	1 · x₁ + -∞
[Begin(x₁)]	=	1 · x₁ + -∞

the rules

b(b(Begin(x0)))	→	Right3(Wait(x0))
End(a(a(b(Right3(x0)))))	→	End(a(a(b(b(b(a(a(Left(x0)))))))))
b(Right2(x0))	→	Right2(Ab(x0))
b(Right3(x0))	→	Right3(Ab(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))
Left(Ab(x0))	→	b(Left(x0))
Left(Aa(x0))	→	a(Left(x0))
Left(Wait(x0))	→	Begin(x0)
b(b(a(a(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

remain.

1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the naturals

[Wait(x₁)]	=	2 · x₁ + 0
[End(x₁)]	=	1 · x₁ + 2
[a(x₁)]	=	1 · x₁ + 0
[Right1(x₁)]	=	8 · x₁ + 0
[Right2(x₁)]	=	2 · x₁ + 9
[Ab(x₁)]	=	2 · x₁ + 0
[b(x₁)]	=	2 · x₁ + 0
[Right4(x₁)]	=	1 · x₁ + 1
[Aa(x₁)]	=	1 · x₁ + 0
[Left(x₁)]	=	2 · x₁ + 0
[Right3(x₁)]	=	8 · x₁ + 0
[Begin(x₁)]	=	4 · x₁ + 0

the rules

b(b(Begin(x0)))	→	Right3(Wait(x0))
End(a(a(b(Right3(x0)))))	→	End(a(a(b(b(b(a(a(Left(x0)))))))))
b(Right3(x0))	→	Right3(Ab(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))
Left(Ab(x0))	→	b(Left(x0))
Left(Aa(x0))	→	a(Left(x0))
Left(Wait(x0))	→	Begin(x0)
b(b(a(a(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

remain.

1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over (3 x 3)-matrices with strict dimension 1 over the naturals

[Wait(x₁)]

1	0	1
0	0	1
0	0	0

· x₁ +

0	0	0
1	0	0
0	0	0

[End(x₁)]

1	0	1
1	0	1
0	0	0

· x₁ +

1	0	0
1	0	0
0	0	0

[a(x₁)]

1	0	0
0	0	0
0	0	1

· x₁ +

0	0	0
0	0	0
0	0	0

[Right1(x₁)]

1	1	1
0	0	0
0	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Right2(x₁)]

1	0	0
0	0	0
1	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Ab(x₁)]

1	0	0
0	1	0
0	1	1

· x₁ +

0	0	0
0	0	0
0	0	0

[b(x₁)]

1	0	0
0	1	0
0	1	1

· x₁ +

0	0	0
0	0	0
0	0	0

[Right4(x₁)]

1	1	0
0	0	0
1	1	0

· x₁ +

0	0	0
0	0	0
0	0	0

[Aa(x₁)]

1	0	0
0	0	0
0	1	1

· x₁ +

0	0	0
0	0	0
0	0	0

[Left(x₁)]

1	0	0
0	1	0
0	1	1

· x₁ +

1	0	0
0	0	0
0	0	0

[Right3(x₁)]

1	0	0
0	1	0
0	1	1

· x₁ +

1	0	0
0	0	0
1	0	0

[Begin(x₁)]

1	0	1
0	0	1
0	0	0

· x₁ +

1	0	0
1	0	0
0	0	0

the rules

b(b(Begin(x0)))	→	Right3(Wait(x0))
b(Right3(x0))	→	Right3(Ab(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))
Left(Ab(x0))	→	b(Left(x0))
Left(Aa(x0))	→	a(Left(x0))
Left(Wait(x0))	→	Begin(x0)
b(b(a(a(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

remain.

1.1.1.1.1.1.1.1.1.1.1 Rule Removal

Using the linear polynomial interpretation over the arctic semiring over the integers

[Wait(x₁)]	=	13 · x₁ + -∞
[a(x₁)]	=	0 · x₁ + -∞
[Right1(x₁)]	=	8 · x₁ + -∞
[Right2(x₁)]	=	8 · x₁ + -∞
[Ab(x₁)]	=	8 · x₁ + -∞
[b(x₁)]	=	8 · x₁ + -∞
[Right4(x₁)]	=	2 · x₁ + -∞
[Aa(x₁)]	=	0 · x₁ + -∞
[Left(x₁)]	=	9 · x₁ + -∞
[Right3(x₁)]	=	10 · x₁ + -∞
[Begin(x₁)]	=	15 · x₁ + -∞

the rules

b(Right3(x0))	→	Right3(Ab(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))
Left(Ab(x0))	→	b(Left(x0))
Left(Aa(x0))	→	a(Left(x0))
b(b(a(a(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

remain.

1.1.1.1.1.1.1.1.1.1.1.1 String Reversal

Since only unary symbols occur, one can reverse all terms and obtains the TRS

Right3(b(x0))	→	Ab(Right3(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))
Ab(Left(x0))	→	Left(b(x0))
Aa(Left(x0))	→	Left(a(x0))
b(a(a(b(b(x0)))))	→	a(a(b(b(b(a(a(x0)))))))

1.1.1.1.1.1.1.1.1.1.1.1.1 Bounds

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

final states:

{15, 13, 11, 9, 8, 6, 4, 1}

transitions:

15 → 17

15 → 12

9 → 10

1 → 3

8 → 3

6 → 7

52 → 19

4 → 5

28 → 18

17 → 21

24 → 45

a₀(19) → 20

a₀(2) → 14

a₀(14) → 16

a₀(20) → 15

Right2₀(2) → 7

Right4₀(2) → 10

Right1₀(2) → 5

a₁(45) → 46

a₁(21) → 22

a₁(27) → 28

a₁(22) → 23

a₁(51) → 52

a₁(26) → 27

a₁(46) → 47

a₁(50) → 51

Aa₀(5) → 4

Aa₀(7) → 6

Aa₀(10) → 9

Aa₀(3) → 8

b₁(23) → 24

b₁(48) → 49

b₁(47) → 48

b₁(24) → 25

b₁(49) → 50

b₁(25) → 26

b₀(2) → 12

b₀(16) → 17

b₀(17) → 18

b₀(18) → 19

Ab₀(3) → 1

Right3₀(2) → 3

f12₀ → 2

Left₀(14) → 13

Left₀(12) → 11

15	→	17
15	→	12
9	→	10
1	→	3
8	→	3
6	→	7
52	→	19
4	→	5
28	→	18
17	→	21
24	→	45
a₀(19)	→	20
a₀(2)	→	14
a₀(14)	→	16
a₀(20)	→	15
Right2₀(2)	→	7
Right4₀(2)	→	10
Right1₀(2)	→	5
a₁(45)	→	46
a₁(21)	→	22
a₁(27)	→	28
a₁(22)	→	23
a₁(51)	→	52
a₁(26)	→	27
a₁(46)	→	47
a₁(50)	→	51
Aa₀(5)	→	4
Aa₀(7)	→	6
Aa₀(10)	→	9
Aa₀(3)	→	8
b₁(23)	→	24
b₁(48)	→	49
b₁(47)	→	48
b₁(24)	→	25
b₁(49)	→	50
b₁(25)	→	26
b₀(2)	→	12
b₀(16)	→	17
b₀(17)	→	18
b₀(18)	→	19
Ab₀(3)	→	1
Right3₀(2)	→	3
f12₀	→	2
Left₀(14)	→	13
Left₀(12)	→	11