“…Now, the proof is straightforward. δ(q 1 , α) = 0.9, r 2 : δ(q 1 , q 4 , λ) = 0.8, r 3 : δ(q 2 , q 4 , λ) = 0.8, r 4 : δ(q 3 , q 4 , λ) = 0.8, r 5 : δ(q 4 , q 2 , λ) = 0.8, r 6 : δ(q 1 , q 1 , q 3 , σ ) = 0.7, r 7 : δ(q 1 , q 2 , q 3 , σ ) = 0.7, r 8 : δ(q 1 , q 3 , q 2 , σ ) = 0.6, r 9 : δ(q 1 , q 4 , q 4 , σ ) = 0.6, r 10 : δ(q 2 , q 1 , q 3 , σ ) = 0.6, r 11 : δ(q 2 , q 2 , q 3 , σ ) = 0.6, r 12 : δ(q 2 , q 3 , q 2 , σ ) = 0.6, r 13 : δ(q 2 , q 4 , q 4 , σ ) = 0.2, r 14 : δ(q 3 , q 1 , q 2 , σ ) = 0.3, r 15 : δ(q 3 , q 2 , q 2 , σ ) = 0.3, r 16 : δ(q 3 , q 3 , q 3 , σ ) = 0.3, r 17 : δ(q 3 , q 4 , q 4 , σ ) = 0.2, r 18 : δ(q 4 , q 1 , q 4 , σ ) = 0.1, r 19 : δ(q 4 , q 2 , q 4 , σ ) = 0.1, r 20 : δ(q 4 , q 3 , q 4 , σ ) = 0.1, r 21 : δ(q 4 , q 4 , q 4 , σ ) = 1}. Now, for each i ∈ {1, 2, 3, 4} we have f * (q i ) = i − 1 and then f * (r 16 M = ( , Q, , δ, , ρ, β) be an FFTA.…”