NO Nontermination Proof

Nontermination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Loop

The following loop proves nontermination.

t0 = a(c(b(c(a(b(a(x0)))))))
→1 a(b(b(a(c(a(b(a(x0))))))))
→ε b(c(a(b(a(c(a(b(a(x0)))))))))
→1.1 b(c(b(c(a(a(c(a(b(a(x0))))))))))
→1.1.1.1.1.1.1 b(c(b(c(a(a(c(b(c(a(a(x0)))))))))))
→1.1.1.1.1.1 b(c(b(c(a(a(b(b(a(c(a(a(x0))))))))))))
→1.1.1.1.1 b(c(b(c(a(b(c(a(b(a(c(a(a(x0)))))))))))))
→1.1.1.1 b(c(b(c(b(c(a(c(a(b(a(c(a(a(x0))))))))))))))
→1.1.1.1.1.1.1.1 b(c(b(c(b(c(a(c(b(c(a(a(c(a(a(x0)))))))))))))))
→1.1.1.1.1.1.1 b(c(b(c(b(c(a(b(b(a(c(a(a(c(a(a(x0))))))))))))))))
→1.1.1.1.1.1 b(c(b(c(b(c(b(c(a(b(a(c(a(a(c(a(a(x0)))))))))))))))))
→1.1.1 b(c(b(b(b(a(c(b(c(a(b(a(c(a(a(c(a(a(x0))))))))))))))))))
= t11
where t11 = C[t0σ] and σ = {x0/c(a(a(c(a(a(x0))))))} and C = b(c(b(b(b(☐)))))