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Is there a proof for the logical equivalence of $a \rightarrow b$ and $\lnot a \vee b$?

Austin Mohr
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4 Answers4

17

Try using a truth table comparing $p\rightarrow q$ with $\lnot p \lor q$.
That's as good a proof as you need.

$p$ | $q\,$ |$\;p\rightarrow q$ | $\lnot p \lor q$
T | T |$\quad$ T $\quad$|$\quad$ T
T | F |$\quad$ F $\quad$|$\quad$ F
F | T |$\quad$ T $\quad$|$\quad$ T
F | F |$\quad$ T $\quad$|$\quad$ T

Both expressions evaluate false if and only if $\,p\,$ is true and $\,q\,$ is false.

For developing a bit of intuition regarding the conditional (logical implication): see, e.g.,
$(1)$ this post, and
$(2)$ this post.

Community
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amWhy
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  • This is basically my problem. I am having hard time understanding imples in logic. Is it like truth of an expression always implies truthness of another? and falseness of p can imply q to be true as well as false? –  Nov 24 '12 at 21:48
  • Yes, exactly. false implies anything: true or false. But for $p \implies q$ to be true when $p$ is true, then $q$ must be true. – amWhy Nov 24 '12 at 21:50
  • Trust me, you're not the only one who finds "implies" (material implication) difficult to comprehend, as you can see by the links I've included above. It's a very common difficulty encountered by students. – amWhy Nov 24 '12 at 22:05
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    Also worth noting is, using De Morgan's laws, the expression for implication is equivalent to $\neg(p\land\neg q)$, which gives us some justification for proof by contradiction. I.e., if $p\land\neg q$ gives us a contradiction, its negation ($p\Longrightarrow q$) must be true. – Alex Nelson Nov 24 '12 at 22:09
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    Note that the truth tables are equivalent because we define the truth table for "implies" in a certain way. – Code-Guru Dec 11 '12 at 04:24
  • As much as I dislike truth tables in general, I would also go for them in this particular case. – Adam Nov 20 '13 at 23:11
  • I agree, @Adam. – amWhy Nov 20 '13 at 23:20
  • "That's as good a proof as you need."

    But a formal proof will tell us that this holds in Bochvar's 3-valued logic as well as some other systems with equivalent axiom sets, which have more than two truth values. The truth-table for Bochvar's 3-valued logic needs more than this answer, or you have to invoke a series of lemmas which ensure that you can decide such questions about those other systems with more truth values by two-valued truth tables. Anyways, nice answer, just a comment.

     – Doug Spoonwood May 15 '14 at 19:41
  • @amWhy The question is why would the logical implication φ → ψ, universally read and understood as "If φ, then ψ", have the same truth values as the material implication φ ⊃ ψ. – Speakpigeon Aug 12 '20 at 07:49
  • @amWhy "Trust me, you're not the only one who finds "implies" (material implication) difficult to comprehend, as you can see by the links I've included above. It's a very common difficulty encountered by students". The material implication is on the contrary very easy to understand. It is on the level of the addition table. What students don't understand, and for very good reasons, is why would the logical implication φ → ψ, universally read and understood as "If φ, then ψ", have the same truth values as the material implication φ ⊃ ψ. – Speakpigeon Aug 12 '20 at 07:53
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Think of it this way. Suppose I signed a contract to you, saying "If I win the lottery, then I will give you $\$1$ billion." The only situation in which my statement would be false (contract violated) is: I win the lottery and don't give you the money I promised. Otherwise, I'm just fine. In particular, if I never win the lottery, you have no claim on any of my money. On the other hand, I can give you the money just for kicks, and I still haven't violated the contract.

Similarly, what if I said "If it's raining, I carry an umbrella." You couldn't call me a liar if it isn't raining, nor if I have an umbrella (regardless of the weather).

A general conditional statement "if (hypothesis) then (conclusion)" is (by this reasoning) the same as saying "(hypothesis) is false or (conclusion) is true."

Cameron Buie
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    "On the other hand, I can give you the money just for kicks, and I still haven't violated the contract." My favourite representation of this is the logician whose son was well-behaved but still went to bed without any supper. – Joe Z. Jan 13 '13 at 14:17
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Suppose we are for the moment using $p \to q$ for the informal indicative conditional, if $p$ then $q$.

  1. Suppose we are given that if $p$ then $q$ (i.e. $p \to q$). Then, since we have $\neg p \lor p$ it follows (arguing by cases) that $\neg p \lor q$.
  2. Suppose, conversely, we are given $\neg p \lor q$ is true. Then if $p$ is true that rules out the first disjunct, so we have $q$. In short, if $p$ then $q$, (i.e. $p \to q$).

That pair of arguments makes it rather plausible that $p \to q$ (the indicative conditional) is inter-dervable with $\neg p \lor q$, and so is equivalent at least as far as truth-conditions go with $\neg p \lor q$.

OK: That gives us some reason to suppose that the truth-functional account of the conditional isn't just fanciful -- it gets something important right. Or so a familiar story goes.

But do note (important point!) it is a further question whether the truth-functional story about 'if' is the whole story.

Compare, arguably '$p$ but $q$' is truth-functionally equivalent to $p \land q$: however 'but' doesn't mean just bare conjunction. 'But' does more than '$\land$' -- very roughly it signals the speaker's sense that the joint truth of $p$ and $q$ is unexpected/surprising. Well likewise, perhaps "if p then q" does more than '$\neg p \lor q$'. E.g. it arguably signals the speaker's willingness to infer $q$ should it turn out that $p$. For more on this, see e.g. http://www.logicmatters.net/resources/pdfs/Conditionals.pdf

Peter Smith
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  • Your number 2 argument is actually (¬p ∨ q) → ((p → (p → q)). Very, very different from (¬p ∨ q) → (p → q). – Speakpigeon Aug 11 '20 at 17:32
  • @Speakpigeon It can't be. What you give is a wff, not an argument. – Peter Smith Aug 12 '20 at 11:50
  • And you don't know how to translate an argument into a wff? – Speakpigeon Aug 12 '20 at 14:05
  • Here is how Wikipedia puts it: "Suppose, conversely, we are given ¬P ∨ Q. Then if P is true that rules out the first disjunct, so we have Q. In short, P → Q. However if P is false, then this entailment fails, because the first disjunct ¬P is true which puts no constraint on the second disjunct Q. Hence, nothing can be said about P → Q. In sum, the equivalence in the case of false P is only conventional, and hence the formal proof of equivalence is only partial". https://en.wikipedia.org/wiki/Material_implication_(rule_of_inference) – Speakpigeon Aug 12 '20 at 14:31
2

I'm gonna use p and q.
Natural deduction: \begin{align} &1\ p \rightarrow q &&P\\ &2\ |\ \lnot (\lnot p \lor q) &&H\\ &3\ |\ |\ p &&H\\ &4\ |\ |\ q &&1,3 I \rightarrow\\ &5\ |\ |\ \lnot p \lor q &&4 I \lor\\ &6\ |\ |\ \lnot ( \lnot p \lor q) \land (\lnot p \lor q) && 2,5 I \lor \bot\\ &7\ |\ \lnot p&&3-6 I \lnot\\ &8\ |\ \lnot p \lor q&&7I\lor\\ &9\ |\ (\lnot p \lor q) \land \lnot (\lnot p \lor q)&&2,8\land \bot\\ &10\ \lnot p \lor q&&2\lnot \end{align}

Hope this helps you!

  • @user94342 You have proved the trivial property (p → q) → (¬p ∨ q), which follows anyway from the definition of p → q as "If p, then q". What really needs to be proven is (¬p ∨ q) → (p → q). Can you do it? – Speakpigeon Aug 12 '20 at 07:24