YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

b(b(a(b(x0))))	→	b(a(b(b(b(b(x0))))))
b(a(b(b(x0))))	→	b(b(b(x0)))
b(b(b(x0)))	→	b(b(a(a(b(a(b(x0)))))))

Proof

1 String Reversal

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

b(a(b(b(x0))))	→	b(b(b(b(a(b(x0))))))
b(b(a(b(x0))))	→	b(b(b(x0)))
b(b(b(x0)))	→	b(a(b(a(a(b(b(x0)))))))

1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.

b^#(a(b(b(x0))))	→	b^#(a(b(x0)))
b^#(a(b(b(x0))))	→	b^#(b(a(b(x0))))
b^#(a(b(b(x0))))	→	b^#(b(b(a(b(x0)))))
b^#(a(b(b(x0))))	→	b^#(b(b(b(a(b(x0))))))
b^#(b(a(b(x0))))	→	b^#(b(x0))
b^#(b(a(b(x0))))	→	b^#(b(b(x0)))
b^#(b(b(x0)))	→	b^#(a(a(b(b(x0)))))
b^#(b(b(x0)))	→	b^#(a(b(a(a(b(b(x0)))))))

1.1.1 Dependency Graph Processor

The dependency pairs are split into 2 components.

The 1^st component contains the pair

b^#(a(b(b(x0))))

→

b^#(a(b(x0)))

1.1.1.1 Reduction Pair Processor

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

[a(x₁)]

0	0	0
2	0	1
0	0	0

· x₁ +

0	0	0
0	0	0
0	0	0

[b(x₁)]

0	1	1
0	0	0
0	1	1

· x₁ +

0	0	0
0	0	0
1	0	0

[b^#(x₁)]

0	1	0
0	0	0
0	0	0

· x₁ +

3	0	0
0	0	0
0	0	0

all pairs could be removed.

1.1.1.1.1 P is empty

There are no pairs anymore.

The 2^nd component contains the pairs

b^#(b(a(b(x0))))	→	b^#(b(b(x0)))
b^#(b(a(b(x0))))	→	b^#(b(x0))

1.1.1.2 Reduction Pair Processor

Using the linear polynomial interpretation over (4 x 4)-matrices with strict dimension 1 over the naturals

[a(x₁)]

0	1	1
1	0	0
0	0	0
0	0	0

· x₁ +

0	0	0	0
0	0	0	0
0	0	0	0
0	0	0	0

[b(x₁)]

0	1	0	0
0	0	0	0
1	0	0	0
0	1	0	0

· x₁ +

0	0	0	0
1	0	0	0
0	0	0	0
0	0	0	0

[b^#(x₁)]

1	0	1	0
0	0	0	0
0	0	0	0
0	0	0	0

· x₁ +

0	0	0	0
0	0	0	0
0	0	0	0
0	0	0	0

the pair

b^#(b(a(b(x0))))

→

b^#(b(b(x0)))

remains.

1.1.1.2.1 Reduction Pair Processor

Using the linear polynomial interpretation over (5 x 5)-matrices with strict dimension 1 over the naturals

[a(x₁)]

0	0	0
1	1	0
0	0	0
0	0	1
0	1	0

· x₁ +

0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0

[b(x₁)]

1	0	0
0	1	1
0	0	0
0	0	0
0	0	0

· x₁ +

0	0	0	0	0
0	0	0	0	0
1	0	0	0	0
0	0	0	0	0
0	0	0	0	0

[b^#(x₁)]

1	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0

· x₁ +

0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0
0	0	0	0	0

all pairs could be removed.

1.1.1.2.1.1 P is empty

There are no pairs anymore.