-1

Here's the issue, there's no usage of derived rules allowed. So no DeMorgan's Law. All that's allowed is the basic TFL elimination/introduction rules, IP, (e)X(plosion), and ⊥. I'm absolutely lost on how to get -Q, any help would be appreciated!

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    Hint: proof by negation, assuming Q first, then Q ∧ P per TFL ∧-intro rule based on all premises, then explode... Commented Feb 13, 2022 at 19:05
  • @DoubleKnot Aye Thank you, this one was holding me up really bad
    – BeepBoop69
    Commented Feb 13, 2022 at 19:22
  • 1
    This proof technique is a common "holding up"... There's some old saying To speak of the false is to reveal the true..., but it's hard to do so for some especially beginners... Commented Feb 13, 2022 at 19:47
  • @DoubleKnot, there is no need to explode. Rather use negation introduction. Commented Feb 17, 2022 at 1:18
  • @GrahamKemp thx for your feedback in the language of logicians which is meant for a formal answer. My above colloquial language was meant as an informal hint... Commented Feb 17, 2022 at 3:31

1 Answer 1

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All that's allowed is the basic TFL elimination/introduction rules, IP, (e)X(plosion), and ⊥. I'm absolutely lost on how to get -Q,

To introduce a negation, we use the rule of negation introduction.

Assume Q, somehow derive a contradiction, and discharge that assumption with the rule.

1|  P
2|_ ¬(Q ∧ P)
3|  |_ Q         H
 |  |  :   
m|  |  ⊥        ¬E ?,?
n|  ¬Q          ¬I 3-m

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