Good afternoon. Consider the following example:

(A → (B C)) Ā C (the first C, too, with a feature on the top, couldn't find that symbol)

The main problem with " → ", I don't know how to break that down. I would be grateful if comments. Thank you.

(A → (B C)) Ā C (the first C, too, with a feature on the top, couldn't find that symbol)

The main problem with " → ", I don't know how to break that down. I would be grateful if comments. Thank you.

asked April 3rd 20 at 18:49

2 answers

answered on April 3rd 20 at 18:51

First, the usual implication

it can be converted to

Then look for the gluing.

`A → (B !C) = (A → B) (A → !C)`

it can be converted to

`(!A B)(!A !C)`

Then look for the gluing.

answered on April 3rd 20 at 18:53

The first part:

(A->(B^C)) == (A->B) ^ (A->!C)

(A->B) ^ (A->!C) == !((!A|B)(A|!B)(A|C)(!A|!C))

!((!A|B)(A|!B)(A|C)(!A|!C)) == !((!A ^ A + !B ^ B)(!A ^ A + !C ^ !C))

!((!A ^ A + !B ^ B)(!A ^ A + !C ^ !C)) == !(0 * !C) == 1

Second part:

1 + !A ^ C == 1 for any A and C

(A->(B^C)) == (A->B) ^ (A->!C)

(A->B) ^ (A->!C) == !((!A|B)(A|!B)(A|C)(!A|!C))

!((!A|B)(A|!B)(A|C)(!A|!C)) == !((!A ^ A + !B ^ B)(!A ^ A + !C ^ !C))

!((!A ^ A + !B ^ B)(!A ^ A + !C ^ !C)) == !(0 * !C) == 1

Second part:

1 + !A ^ C == 1 for any A and C

You can also draw the truth table and verify commented on April 3rd 20 at 18:56

Find more questions by tags AlgebraFuzzy logic

(A -> (B !C)) !A C

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(!A B !C) !A C

But then, no how.... - eloise.Lang commented on April 3rd 20 at 18:54