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

Proof

1 Loop

The following loop proves nontermination.

t0 = b(a(a(a(b(b(c(c(a(x0)))))))))
→1.1 b(a(b(a(b(b(c(c(a(x0)))))))))
→1.1.1.1 b(a(b(a(c(c(a(a(a(c(a(x0)))))))))))
→1.1.1.1.1.1.1 b(a(b(a(c(c(a(b(a(c(a(x0)))))))))))
→1.1.1.1.1.1.1.1 b(a(b(a(c(c(a(b(b(c(a(x0)))))))))))
→1.1.1 b(a(b(b(c(c(a(b(b(c(a(x0)))))))))))
→1.1.1.1 b(a(b(b(c(a(b(b(c(a(x0))))))))))
→1.1.1.1.1.1 b(a(b(b(c(a(c(c(a(a(a(a(x0))))))))))))
→1.1.1.1.1.1.1.1 b(a(b(b(c(a(c(c(a(a(a(x0)))))))))))
→1.1 b(a(c(c(a(a(a(a(c(c(a(a(a(x0)))))))))))))
→1.1.1.1.1.1 b(a(c(c(a(a(b(a(c(c(a(a(a(x0)))))))))))))
→1.1.1.1.1.1.1 b(a(c(c(a(a(b(b(c(c(a(a(a(x0)))))))))))))
→1.1 b(a(c(a(a(b(b(c(c(a(a(a(x0))))))))))))
→1.1 b(a(a(a(b(b(c(c(a(a(a(x0)))))))))))
= t13
where t13 = t0σ and σ = {x0/a(a(x0))}