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I tried to solve this deduction but I got stuck:

assumed $\phi ∨ \psi$

assumed $¬\phi$

what can I do now?

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    Use $\lor\text{-Elim}$ on $\phi \lor \psi$. – 2017-01-02

3 Answers 3

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I don't know if the system you use has an explicit contradiction symbol and related inference rules, but if it does, this might work:

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With natural deduction:

(1) $\phi \lor \psi \qquad$ Hyp.

(2) $\lnot \phi \qquad$ Hyp.

(3) $\phi \lor \psi\qquad $ Rep (1)

(4) $\phi \qquad$ Hyp.

(5) $\lnot \psi \qquad$ Hyp.

(6) $\phi \qquad$ Hep. (4)

(7) $\lnot \phi \qquad$ Rep. (2)

(8) $\phi \land \lnot \phi \qquad \land$ Intro. (6),(7)

(9) $\lnot \lnot \psi \qquad \lnot$ Intro. (5)-(8)

(10) $\psi \qquad \lnot$ Elim. (9)

(11) $\phi \rightarrow \psi \qquad \rightarrow$ Intro. (4)-(10)

(12) $\psi \qquad$ Hyp.

(13) $\psi \qquad$ Rep. (12)

(14) $\psi \rightarrow \psi \qquad \rightarrow$ Intro. (12)-(13)

(15) $\psi \qquad \lor$ Elim. (3),(11),(14)

(16) $\lnot \phi \rightarrow \psi \qquad \rightarrow$ Intro. (2)-(15)

(17) $\left(\phi \lor \psi\right) \rightarrow \left(\lnot \phi \rightarrow \psi \right) \qquad \rightarrow$ Intro. (1)-(16)

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    $\lor$-elim does not work that way: you hace to use both disjuncts. – 2017-01-03
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    @MauroALLEGRANZA Hey thanks for your comment, I didn't even notice it. Now it's fixed. – 2017-01-03
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If A or B:

  If not A:

    A or B.   [(1) You need to use this and the only way is disjunction elimination.]

    If A:

      ?   [(3) You may need principle of explosion or proof by contradiction here.]

    If B:

      ?

    B.   [(2) This is your goal so you should aim to deduce this in both of the above cases.]

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    Disjunction elimination isn't necessary. We could have a rule of inference that says "from ($\phi$$\lor$$\psi$) you may infer (($\phi$$\rightarrow$$\psi$)$\rightarrow$$\psi$)." So, from (A$\lor$B) we have ((A$\rightarrow$B)$\rightarrow$B). Then assuming $\lnot$A we can get to (A$\rightarrow$B). Then by detachment, we have B. So, ($\lnot$A$\rightarrow$B). – 2017-01-05
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    @DougSpoonwood: Thank you for the sensible comment. You are right that you **could have** a rule of inference as you say. However, when I said "disjunction elimination" I literally meant any inference rule that eliminates at least one disjunction, and that would include your proposed rule. That is what is often meant when we say that we need introduction and elimination rules for each symbol. So my post is not incorrect in this sense. For the sake of beginners, I think it is best to stick with the common kind of inference rules and leave out some of the minute technical details. – 2017-01-05
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    My comment works for the rule of inference of $\lor$ Elim as used in Bram28's answer. I see your point about the rule I had above as a disjunction elimination rule. Sure, in that sense it works as a disjunction elimination rule. But, I don't know why anyone says that we need introduction and elimination rules for each symbol. Jaskowksi's original paper on natural deduction doesn't have an introduction rule for 'N'. It only has an elimination rule for 'N'. Mendelson's axiom set for propositional calculus along with the deduction meta-theorem imply that no introduction for negation is needed. – 2017-01-05
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    To make that more concrete, conditional introduction, conditional elimination: {(cond p q), p} $\vdash$ q, and strong negation elimination: {(cond (neg p) q), (cond (neg p) (neg q))} $\vdash$ p allow you to derive every single negation introduction rule. So, no negation introduction rule is necessary for some natural deduction system, and thus we don't need introduction and elimination rules for each symbol. – 2017-01-05
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    @DougSpoonwood: Every proof of the tautology "$\neg ( A \land \neg A )$" must use an inference rule that permits deduction of a negated statement. In the system that you mention, $\neg$-elim can act as a negation introduction rule. Mendelson's has MP, which can introduce any symbol by using the appropriate axiom, so it doesn't count either. Just like Gentzen, I prefer systems that have a clear correspondence between rules and semantics, thus I prefer explicit rather than implicit negation introduction. And please use standard notation or ASCII (as in your first comment) or English, thanks. – 2017-01-06