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I am having some trouble understanding how I go about putting the following expression into a truth table form, mainly because the first part is confusing me. I was hoping to get some input and help.

F = ~(A1 --> A2) ∨ (A1∧A2) ∨ (A2∧A3)

I realise I have three variables and I know I have to calculate each case, which is 2^3, so 8 sets. I am having trouble with the first part and have no idea how to begin calculating the truth in each case of ~(A1-->A2).

Any help would be greatly appreciated.

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    You need the three initial columns for the variables $A_1,A_2,A_3$ and (very useful) three "intermediate columns for the sub-formulas : $\lnot (A_1 \to A_2), (A_1 \land A_2), (A_2 \land A_3)$ plus the final column for the complete formula. – 2017-01-18
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    The "intermediate" columns must be computed according to the truth-tables for the connectives. For the first-one, use $\to$ and then reverse all **T** with **F** and vice versa: this is the effect of the leading negation sign. – 2017-01-18
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    Hi Mauro, thanks for the help. This makes sense. One question: does the ~ apply to all brackets or just the first one? From your posts I understand that I need only to apply the negation sign to the first intermediate column and simply work out the → and then reverse it, due to that negation. Is that right? I don't need to reverse the other intermediate columns? – 2017-01-18
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    Correct; the negation sign would apply to the complete formula only if written : $\lnot [(A_1 \to A_2) \lor \ldots \lor \ldots]$. – 2017-01-18
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    Greatly appreciated! I think I can now happily progress with this question! Thank you. – 2017-01-18

1 Answers 1

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 A1 | A2 | A3 | ¬(A1⟹A2) | A1∧A2 | A2∧A3 | F
----|----|----|-----------|-------|-------|---
 0  | 0  | 0  |     0     |   0   |   0   | 0
 0  | 0  | 1  |     0     |   0   |   0   | 0
 0  | 1  | 0  |     0     |   0   |   0   | 0
 0  | 1  | 1  |     0     |   0   |   1   | 1
 1  | 0  | 0  |     1     |   0   |   0   | 1
 1  | 0  | 1  |     1     |   0   |   0   | 1
 1  | 1  | 0  |     0     |   1   |   0   | 1
 1  | 1  | 1  |     0     |   1   |   1   | 1