NO Nontermination Proof

Nontermination Proof

by ttt2 (version ttt2 1.15)

Input

The rewrite relation of the following TRS is considered.

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

Proof

1 String Reversal

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

1.1 Loop

The following loop proves nontermination.

t0 = b(a(b(a(b(c(x0))))))
→1.1.1.1 b(a(b(a(c(x0)))))
→1.1 b(a(a(a(b(c(a(b(c(x0)))))))))
→1.1.1.1.1.1 b(a(a(a(b(c(b(c(x0))))))))
→ε a(a(b(c(a(b(a(a(b(c(b(c(x0))))))))))))
→1.1.1.1.1.1.1.1.1.1 a(a(b(c(a(b(a(a(b(c(c(x0)))))))))))
→1.1.1.1 a(a(b(c(b(a(a(b(c(c(x0))))))))))
→1.1.1.1 a(a(b(c(a(a(b(c(a(b(a(b(c(c(x0))))))))))))))
→1.1.1.1.1.1 a(a(b(c(a(a(c(a(b(a(b(c(c(x0)))))))))))))
→1.1.1.1 a(a(b(c(a(c(a(b(a(b(c(c(x0))))))))))))
→1.1.1.1 a(a(b(c(c(a(b(a(b(c(c(x0)))))))))))
→1.1.1 a(a(b(a(b(a(b(c(c(x0)))))))))
= t11
where t11 = C[t0σ] and σ = {x0/c(x0)} and C = a(a(☐))