How can I prove this true:
$$p \vee q, \neg q \vdash q$$
Using natural deduction?
I have tried to eliminate the disjunction but didn't go anywhere and working with proof by contradiction didn't help me either..
Knowing one term is false, how can I prove the other is true in a disjuntion?
1
$\begingroup$
logic
propositional-calculus
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4You won't be able to prove it because it's not true. Is there a typo? – 2017-02-15
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4Expanding on Git Gud's comment, what is true is $ p \vee q, \neg q \vdash p$. But what you've written is not true. – 2017-02-15
1 Answers
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Using the following proof tree:
$\cfrac{\cfrac{p\vee q\quad \cfrac{{}^\dagger p^1 \quad \neg p}{\bot}\,\neg E \quad \cfrac{{}^\dagger q^2 \quad {}^\dagger (\neg q)^3}{\bot} \, \neg E}{\bot} \,\vee E,1,2}{q} \,\bot E,3$
${}^\dagger$ denotes marked assumptions, $\neg E$ is negation elimination, $\vee E$ is or elimination and $\bot E$ is absurdity elimination. Notation is from Sets, Models and Proofs