Hi I have been trying this question for a while but I'm unable to construct a proof for the following question, these rules I have proved and these can be used in the proof any help would be much appreciated
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1It is good to show what you have tried – FShrike Aug 21 '21 at 08:53
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What does DM stand for? – FShrike Aug 21 '21 at 08:53
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@FShrike From the looks of it, it looks like DM stands for DeMorgan's law – Constantly confused Aug 21 '21 at 10:46
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1Please don't change the question after people answered it. That's a rude way to thank them for their time and efforts. – Taladris Aug 23 '21 at 15:12
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Using the "additional" rules (that are not part of the standard ND set of rules) we have:
$(p \land q) \to r$ --- premise
$\lnot [(p \to r) \lor (q \to r)]$ - assumed [a]
$\lnot (p \to r) \land \lnot (q \to r)$ --- from 2) by (DM)
$\lnot (p \to r)$ --- from 3)
$(p \land \lnot r)$ --- from 4) by (NoI)
$p$ --- from 5)
$\lnot r$ --- from 5)
$\lnot (q \to r)$ --- from 3)
$(q \land \lnot r)$ --- from 8) by (NoI)
$q$ --- from 9)
$(p \land q)$ --- from 6) and 10)
$r$ --- from 1) and 11)
$\bot$ --- Contradiction, from 7) and 12)
- $(p \to r) \lor (q \to r)$ --- from 2) and 13) by Double Negation, discharging [a]
Mauro ALLEGRANZA
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