Current Knowledge:

  • Know off by heart the standard rules + variant rules for the inference system (natural deduction).

  • Know about predicate logic and proportional logic.

Does anyone have a technique in how to approach natural deduction questions, because honestly I can't rack my mind around it. I know practice will help me get better an all, and I have been doing tons of practice. But ever time a new question comes up I just get completely stuck....

If someone can give me some tips/heuristics or something to help me, that would be awesome. Something like what to watch out for if the conclusion is supposed to be this, then approach it in this manner...etc . Basic mathematical language type explanations would be preferable.

I have been having trouble with the following example, as of late. If you could base your answers on this,that would be awesome.


¬(P∧ ¬Q) , R ⊢ P → (Q ∧ R)

Side question:

  • Could someone tell me the distinct difference between ⊢ and ⊨ ?

  • What is meant by soundness and completeness ? if you could link this too both predicate logic and proportional logic, that would be awesome.

Again basic mathematical language type explanations would be preferable.

Thanks in advance.

  • $\begingroup$ Side question : see what is the difference between $\vdash$ and $\vDash$ ?. $\endgroup$ – Mauro ALLEGRANZA Jan 2 '17 at 21:09
  • $\begingroup$ See Wiki for completenss. $\endgroup$ – Mauro ALLEGRANZA Jan 2 '17 at 21:33
  • $\begingroup$ $\lnot (P \land \lnot Q) \vdash P \Rightarrow (Q \land R)$ is not a theorem of natural deduction. One quick simplification you can make is that the above deduction is only provable if $\lnot (P \land \lnot Q), P \vdash Q$ is provable (the $R$ is just saying "if I start with $R$ then I can conditionally construct $R$" which is just a weakening). A quick BFS shows that you can't derive anything interesting from $\lnot (P \land \lnot Q), P$. $\endgroup$ – DanielV Jan 3 '17 at 0:18

In order to apply Natural Deduction to the example, we have to start from the sought conclusion : $P \to (Q \land R)$.

It is a conditional, so we have to assume $P$ and try to derive $Q \land R$.

We have $R$ but not $Q$; thus, we need a way to derive it: if we assume $\lnot Q$ and are able to derive a contradiction, we can use double negation elimination to get $Q$.

Thus, we need a further assumption : $P$.

1) $¬(P∧ ¬Q)$ --- 1st premise

2) $R$ --- 2nd premise

3) $P$ --- "temporary" assumption, starting sub-proof [a]

4) $\lnot Q$ --- "temporary" assumption, starting sub-proof [b]

5) $P \land \lnot Q$ --- $\land$-intro, from 3) and 4)

6) $\bot$ --- from 1) and 5), by $\lnot$-elim

7) $Q$ --- from 4) and 6) by double negation elimination, closing sub-proof [b]

8) $Q \land R$ --- $\land$-intro

9) $P \to (Q \land R)$ --- $\to$-intro, from 3) and 8), closing sub-proof [a].

The above derivation amounts to showing that:

$¬(P∧ ¬Q), R \vdash P \to (Q \land R)$

i.e. that $P \to (Q \land R)$ is derivable (in the calculus of Natural Deduction) from the set of premises: $\{ ¬(P∧ ¬Q), R \}$.

In general, to say that:

$\Gamma \vDash \varphi$,

where $\Gamma$ is a set of formulas and $\varphi$ a formula, means that the conclusion $\varphi$ is a logical consequence of the premises.

The calculus is (stongly) complete when :

if $\Gamma \vDash \varphi$, then $\Gamma \vdash \varphi$.

The calculus is sound when :

if $\Gamma \vdash \varphi$, then $\Gamma \vDash \varphi$.

  • $\begingroup$ Unfamiliar with lines 3 and 4. Also, I am limited to only using the standard rules and variant rules. $\endgroup$ – Jack Roberts Jan 2 '17 at 21:40
  • $\begingroup$ @JackRoberts: Every natural deduction system has a mechanism for creating subcontexts in which some assumption is true. That is all that Mauro is doing in lines 3,4. If you don't get it then you'd need to provide your rules. $\endgroup$ – user21820 Jan 3 '17 at 5:45
  • 1
    $\begingroup$ Mauro, I wonder if your answers would be much clearer if you used indentation like in Fitch-style, so that it is clear which assumptions are currently in effect at each line. On Math SE, I use   to make a 'tab'-like space. $\endgroup$ – user21820 Jan 3 '17 at 5:47
  • $\begingroup$ The rules I am restricted too are the fundamental rules and variant rules used in natural deduction , inference system. But ok I will add the rules I am restricted too onto the main question. $\endgroup$ – Jack Roberts Jan 3 '17 at 7:55
  • $\begingroup$ @JackRoberts - the "fundamental" rules manages assumptions, like e.g. $\to$-intro. $\endgroup$ – Mauro ALLEGRANZA Jan 3 '17 at 8:40

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