0
$\begingroup$

Someone sent me this, but since it's been ages since I followed discrete mathematics, it's all very vague to me. The question is to find the errors in this proof:

  1. $\neg p \vee (q \& r)$ (Prem)
  2. $\neg p \rightarrow s$ (Prem)

  3. $\neg p \vee q $ (1. Sim)

  4. $\neg(s \& t)$ (Hyp)

  5. $\neg s$ (4. NC)
  6. $p$ (2, 5; MP)

  7. $q$ (3, 6; DS)

  8. $\neg(s\&t) \rightarrow q$ (4, 7; VB)

  • 1
    The only incorrect deduction seems to be from 4 to 5; from $\neg (s \land t)$ it doesn't follow that $\neg s$ (namely, $s$ could hold while $t$ is false). Other than that there are not so much errors in the proof, but many steps seem rather quick; I've never seen the rule you call 'Sim', for instance, but would rather have a subproof for that. – 2017-01-15
  • 0
    Thanks for your response, I'm going to pass it through. Sim means that line number 3 is a simplification from line number 1, which I was actually doubting could be true after running it through Maple. – 2017-01-15
  • 0
    Mees's comment is right: the conclusion is not implied by the premises. Try with a truth assignment $v$ such that : $v(q)=v(p)=v(t)=$ **f** and $v(s)=$ **t**. – 2017-01-15

2 Answers 2

0

For step 3: I assume $Sim$ stands for simplification, which is typically defined as:

$P \land Q$

$\therefore P$ (or $Q$)

Now, you can only apply inference rule when the statement as a whole is of the form as indicated. that is, you are not allowed to apply these rules to component statements, otherwise I might end up doing something like this:

$(P \land Q) \rightarrow R$

$\therefore P \rightarrow R$ (Sim)

but this is NOT a valid inference! So, step 3 is not a correct application of Sim! Put differently: even though the inferred statement in step 3 does validly follow, this was really just a 'lucky' event. For a proof this step is unacceptable.

For step 6: I assume MP stands for Modus Ponens, which is typically defined as:

$P \rightarrow Q$

$P$

$\therefore Q$

Clearly, step 6 is not following this pattern, so it is not a correct application of MP. Even though, yet again, the inferred statement does happen to validly follow. in particular, step 6 is much closer to an application of Modus Tollens, typically defined as:

$P \rightarrow Q$

$\neg Q$

$\therefore \neg P$

Step 5: i have never seen NC ... But I suspect it stands for Negation Conjunction. As such, the only reasonable inference would be one instance of DeMorgan:

$\neg (P \land Q)$

$\therefore \neg P \lor \neg Q$

but clearly 5 does not follow that pattern. in fact, step 5 flatly does not logically follow.

Step 7: DS undoubtedly stands for Disjunctive syllogism, which is typically defined as:

$P \lor Q$

$\neg P$

$\therefore Q$

Now, if you are real picky, you could complain that step 7 isn't exactly of this form, and that you would first need to double negate the $p$ to $\neg \neg p$ before being able to apply this rule ... Nitpicky!!

0

There's more than the one mistake listed by Mees.

From premise $(1)\to (3)$, we have $\lnot p \lor (q\land r) \equiv (\lnot p \lor q) \land (\lnot p \lor r)\quad {1(i)}$, from which you can then use simplification ($3)$ to infer $\lnot p \lor q \equiv p \rightarrow q$. (Ditto for inferring $\lnot p \lor r \equiv p\rightarrow r)$ . You've also made a mistake immediately after listing the hypothesis $¬(s\land t)$ when you use it to justify $\lnot s$. $$\lnot(s \land t) \equiv \lnot s \lor \lnot t \tag{DeMorgan's Law on $(4)$} $$

From which it is not valid to infer $\lnot s$.

If $\lnot s$ were correct, and given the premise $\lnot p \rightarrow s$, you'd get therefore $\lnot \lnot p \equiv p$ by Modus Tollens, and double negation.

Note: if you change your hypothesis to $\lnot S \land \lnot T$, (or it's equivalent $\lnot (s\lor t)$), then your proof would would follow and but with the conclusion $\therefore (\lnot s \land \lnot t)\rightarrow q$