YES Termination Proof

Termination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Rule Removal

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

1.1.1 String Reversal

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

1.1.1.1 Dependency Pair Transformation

The following set of initial dependency pairs has been identified.
b#(b(x0)) → b#(a(b(x0)))
b#(b(a(b(x0)))) → b#(b(x0))
b#(b(a(b(x0)))) → b#(a(a(b(b(x0)))))
b#(b(a(b(x0)))) → b#(a(b(a(a(b(b(x0)))))))
b#(a(a(b(a(b(x0)))))) → b#(b(x0))

1.1.1.1.1 Dependency Graph Processor

The dependency pairs are split into 1 component.