So, how can I have to prove this using natural deduction: $\lnot p, p \lor q \vdash q$

What I did is:

  1. $\lnot p$
  2. $p \lor q$
  3. p assumption
  4. $\bot$ from 1&3
  5. q from 4

Is it ok 100% ? What can I do to make it perfect ? Thanks!


No, it is not.

You have a disjunction as 2nd premise : thus you have to consider both disjuncts with $(\lor \text E)$.

The first sub-case, with $p$ as assumption, is Ok.

You have to add the second sub-case, with $q$ as assumption, in which case the conclusion $q$ is immediate.

Then, having derived $q$ in both cases, you can use $(\lor \text E)$ and conclude.

The flaw in your derivation is that you have the undischarged assumption 3. Thus, what your derivation amounts to is really :

$¬p, p∨q, p \vdash q$.

  • $\begingroup$ So if I go like this: ... 6. q assumption 7. q from 5&6 is ok ? $\endgroup$ Aug 29 '19 at 8:54
  • $\begingroup$ @RazvanStatescu - correct. $\endgroup$ Aug 29 '19 at 9:01

It helps to use a proof checker to make sure one uses the rules correctly. Here is a proof:

enter image description here

The first five lines are the same as your proof. However, they only considered the left side, $P$, of the disjunction on line 2. You have to also consider the right side, $Q$. Note how that was done in this proof checker simply by stating the assumption on line 6. Line 6 was also the derivation of the goal.

The justification for line 7 was given as "∨E 2, 3–5, 6–6". That can be understood as using disjunction elimination (∨E) on the disjunction on line 2 noted as "2" in the justification. One side of the disjunction started with an assumption on line 3 and derived the goal, $Q$, on line 5. That subproof was noted as "3-5". The other side of the disjunction started with an assumption on line 6 and since it already was what I wanted to derive I ended the subproof on line 6 as well. This was noted as "6-6".

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/


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