7/18/2019 Twist Representation

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Let us prove that  I (B ) =  I (Tw(A, ∇(B ), ∆(B ))).

The inclusion I (B ) ⊆  I (Tw(A, ∇(B ), ∆(B ))) is obvious. Indeed, if (a, b) ∈  I (B ), thena ≥  b,  a  ∈ ∇(B ) and  b  ∈  ∆(B ). In this case (a, b)  ∈  Tw(A, ∇(B ), ∆(B )) and, moreover(a, b) ∈  I (Tw(A, ∇(B ), ∆(B ))) in view of  a ≥  b.

Let (a, b)  ∈   I (Tw(A, ∇(B ), ∆(B ))), then again  a  ≥   b,   a  ∈ ∇(B ) and   b  ∈  ∆(B ), andwe have to prove that (a, b) ∈ B .In view of ∆(B ) =  {d  ∈ A | (1, d)  ∈ B}  we have (1, b)  ∈ B . Moreover, (a, c)  ∈ B   for

some  c ∈ A with  a ≥  c  and (1, c) ∈ B . We have also (1, b ∧ c) ∈ B . Further,

(a, c) →  (b ∧ c, 1) = (a →  (b ∧ c), a) ∈ B .

(a →  (b ∧ c), a) ∧ (c, a) = ((a →  (b ∧ c)) ∧ c, a) = ((a →  (b ∧ c)) ∧ (a ∧ c), a) =

= (((a →  (b ∧ c)) ∧ a) ∧ c, a) = (a ∧ (b ∧ c) ∧ c, a) = (b ∧ c, a) ∈ B .

So we have (a, b ∧ c) ∈ B .

(a, b ∧ c) ∧ (1, b) = (a, (b ∧ c) ∨ b) = (a, b) ∈ B .

Note that the situation becomes more clear, if we consider twist-structures over Heyt-ing algebras, in which case we have 0 in the underlying structure and (0, 1) in everytwist-structure. We can prove that

∇(B ) = {a ∈ A | (a, 0) ∈ B}.

Indeed, if  a ∈ ∇(B ), then (a, c) ∈ B   for some  c ∈ A  with  a ≥  c  and (1, c) ∈ B . We have(a, c) →  (0, 1) = (a →  0, a) ∈ B , whence

(a →  0, a) ∧ (c, a) = ((a →  0) ∧ c, a) = (0, a) ∈ B .

Now let  a ∈ ∇(B ) and  b ∈  ∆(B ), then we have (a, 0), (1, b) ∈ B , consequently,

(a, 0) ∧ (1, b) = (a, b) ∈ B .

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