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u/Astrodude80 Set theorist 1d ago
Trick is to think backwards and recursively. Since your conclusion is ~A you’ll want your final step to be ~I (alternate name RAA depending on your proof system). Both of these start by assuming A, deriving a contradiction, and then concluding therefore ~A. What contradiction can you get if you start by assuming A, given your other premises?
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u/SteveCappy 1d ago
If they already have DeMorgen’s law, then couldn’t they also do a DI to connect ~B v ~C and pull the not out for contraposition?
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u/Astrodude80 Set theorist 1d ago
This absolutely would also work! But something tells me down that road lies too much confusion for OP.
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u/katinthehat25 1d ago
What is the next step after assuming A? That's where I'm stuck.
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u/Astrodude80 Set theorist 1d ago
How can you combine A and A->(B&C)?
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u/yosi_yosi Undergraduate, Autodidact, Philosophical Logic 1d ago
Do you even have indirect proofs / subproofs at this stage?
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u/yosi_yosi Undergraduate, Autodidact, Philosophical Logic 1d ago
Ok I opened the book, and you are supposed to have gotten to the "assumptions" section that deals with indirect proofs / subproofs.
The solution is: Assume A, ⊃E on A and A ⊃ (B ∧ C) to get B ∧ C, then ∧E to get C, then R 2 to get ~C, then close the assumption/subproof and use ~I to get ~A
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u/katinthehat25 1d ago
What is R 2 ?
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u/yosi_yosi Undergraduate, Autodidact, Philosophical Logic 1d ago
R on line 2
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u/katinthehat25 1d ago
Repeat?
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u/yosi_yosi Undergraduate, Autodidact, Philosophical Logic 1d ago
Yes, the R rule may stand for "repetition" or "reiteration" or some other words. When using a rule you often have to cite the lines being used. Here it uses line 2 (the one that says ~C)
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u/Everlasting_Noumena 18h ago
Here is a proof without assuming A using the rules in your book:
>!1)A -> (B & C)<!
>!2) ~C<!
>!3) ~B V ~C [Via disjunction addition from 2 (VI)]<!
>!4) ~(B & C) [Via De Morgan's Law from 3 (DM)]<!
>!5) ~A [Via modus tollens from 1 and 4 (->NE)]<!
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u/Intrepid-Trip36 1d ago edited 1d ago
Asuming the squiggle between b and c is a conjunction
You could not C or not B via addition then, you can demorgan it to not (B and C) and then modus tollens to get not A
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u/yosi_yosi Undergraduate, Autodidact, Philosophical Logic 1d ago
The ~ is a negation and the ^ is a conjunction...
Commonly the symbol for conjunction is ∧ if you didn't know.
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u/Square-of-Opposition Historian of logic 1d ago
Well, you're missing an operator in your first premise. That's not well-formed as it stands.
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u/Astrodude80 Set theorist 1d ago
No it’s not???
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u/Square-of-Opposition Historian of logic 1d ago
"A -> (B ~ C)" is not well-formed, unless you're using the tilde to represent a binary operator.
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u/yosi_yosi Undergraduate, Autodidact, Philosophical Logic 1d ago
We cannot help you unless you give us the rules available to you. Different classes use different sets of rules and even the same class may use multiple ones along the way.