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So far I got this:

$[(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r]$:

  • $p∧ T ∧ [(¬q∧p) ∨ r]$
  • $p∧ [(p∧¬q) ∨ r]$
  • $p \lor r$

$(p ∧ ¬q) ∨ (r ∧ q):$

  • $(p ∧ ¬q) ∨ (q ∧ r)$

  • $(p ∧( ¬q ∨ q)∧ r$

  • $p ∧ T ∧ r$

  • $p ∧ r$

amWhy
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Derek Long
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    By commutativity of $\land$, your first formula is equivalent to $[(p\land\neg q)\lor q]\land[(p\land\neg q)\lor r]$ and your second formula is equivalent to $(p\land\neg q)\lor(q\land r)$. And these are equivalent by the distributive law. – Andreas Blass Jan 24 '19 at 00:11
  • can you please explain a little more on how [(p ∧ ¬q) ∨ q] ∧ [(¬q ∧ p) ∨ r] =(p ∧ ¬q) ∨ (r ∧ q) – Derek Long Jan 24 '19 at 01:22
  • [(p∧¬q) V q ] ∧(p∧¬q) v r ] under first formula? – Derek Long Jan 24 '19 at 01:30
  • Unfortunately, I don't see a way to explain more. My comment already broke the argument into three steps, each applying only one of the laws of propositional logic --- two uses of the commutative law of $\land$ and one use of the distributive law. I see no way to break it down any further. – Andreas Blass Jan 24 '19 at 11:30
  • Thank you! its making a lot more sense! – Derek Long Jan 24 '19 at 16:29

1 Answers1

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You're not doing this right. You can't just move parentheses around like you do.

For example, for the first one, you go from $(p \land \neg q) \lor q$ to $p \land \top$, but that must mean you went from $(p \land \neg q) \lor q$ to $p \land (\neg q \lor q)$ .... which is not right:

In general, $(p \land q) \lor r$ is not equivalent to $p \land (q \lor r)$

You are similarly moving parentheses for the second expression in a way that is not allowed. Think about it: what if this was an algebraic expression using numbers? It would be like going from $(3 + 4) \cdot (5+6)$ to $3 + 4 \cdot 5 + 6$ ... that's clearly not something you can do!

Bram28
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