Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm having great difficulty solving the following problem, and even figuring out where to start with the proof. $$ \neg A\lor\neg(\neg B\land(\neg A\lor B)) $$ Please see the following examples of how to do proofs, I would appreciate it if you could attempt to give me guidance using the tools and the line numbers that it cites similar to those below:

This is a sample proof:

This is a sample proof

This is another sample proof (law of excluded middles):

enter image description here

share|improve this question
What logic are you using ? –  mercio Apr 3 '11 at 10:04
@chandok, This is First order logic. And it is about formal proofs and boolean logic. –  KerxPhilo Apr 3 '11 at 11:10
add comment

3 Answers 3

up vote 3 down vote accepted

first of all the statement is true (checking truth values for $A$ and $B$), even though ive now given two incorrect answers... here we go again!

$\begin{align} & \neg A \lor \neg (\neg B \land (\neg A \lor B))\\ & =\neg A \lor \neg [(\neg B \land \neg A) \lor (\neg B \land B)]\\ & =\neg A \lor \neg [(\neg B \land \neg A) \lor FALSE]\\ & =\neg A \lor \neg(\neg B\land\neg A)\\ & =\neg A\lor(B\lor A) \text{ using } \neg(X\land Y)=\neg X\lor\neg Y\\ & =TRUE \end{align}$

formalize this in whatever system youre using, you might have to prove a few things beforehand...

share|improve this answer
Could you please elaborate further. I'm sorry, but I am having a very hard time comprehending your answer. Thank you so much for your support. –  KerxPhilo Apr 3 '11 at 8:01
How do you get $TRUE$ from $\lnot[(\lnot B\land\lnot A)\lor FALSE]$ ? –  Apostolos Apr 4 '11 at 12:20
add comment

Seems like assuming the negation would give you a contradiction:

use ~ for 'negation', and & for 'and'; then your negated assumption becomes:

A&(~B&(~A\/B)) (Assumption)

From which you get:

~B&(~A\/B) (By & elimination)

I think you can show from here that your (negated) assumption leads to a contradiction, by arguing by cases, from ~A\/B, and showing neither case is possible.

Now, ~A\/B follows using a second & elimination. Show neither ~A nor B is possible from the premises in your negated statement. Then from ~A\/B and ~(~A) and ~B, a contradiction to your negated assumption follows.

Edit: this may be much simpler, tho yoyo's answer may be better than mine in that he gives a direct proof, and mine is by contradiction:

Assume the negation of your statement:

i)A& (~B&(~A\/B))

ii)Conclude A, by detachment.

iii)Conclude ~A by detachment inside of parenthesis.

iv)Negation of 'i)' follows by contradiction A&~A

Where detachment--more precisely, &-detachment-- is the rule that allows us to conclude either A, or B, from a statement A&B. To show it is a valid rule, you can either use a truth table to show it is a tautology/logical truth, derive it from the empty set of premises (this is the definition of theorem I am more familiar with), or,equivalently, show that the negation of either of :(A&B-->A) or of (A&B-->A), is a contradiction.For the second approach, assume A&B, and just conclude A (seems tautological, but it works; many of these arguments in sentence logic seem tautological anyway), or transform the implication A&B->A into the equivalent statement ~(A&B)\/A == ~A\/~B\/A== ~A\/A\/B==(~A\/A)\/B (I'm using here the result that A->B is truth-functionally equiv. to ~A\/B), which is a tautology.

share|improve this answer
What do you mean by detachment? –  KerxPhilo Apr 3 '11 at 8:07
KerxPhilo: I mean that from an expression A&B, we can "detach" either A or B. More precisely, I mean –  gary Apr 3 '11 at 23:36
erxPhilo: I mean that from an expression A&B, we can "detach" either A or B. More precisely, that: i) A&B->A , and ii)A&B-> B, In my proof, specifically, from A&(~B&(~A\/B), we can "detach" either the 'A' on the left, or the expression '(~B&(~A\/B))' to the right of the '&'. Since we have another '&' in the parenthesis '(~B&(~A\/B))', we can detach either '~B'or '~A\ /B' –  gary Apr 3 '11 at 23:48
Sorry for the choppy formatting in the above comments, KerxPhilo. I edited my answer to incorporate everything into the answer. –  gary Apr 4 '11 at 3:28
add comment

The "complicated" formula :


can be re-written, due to the equivalence between $P \rightarrow Q$ and $\lnot P \lor Q$, as :

$A \rightarrow \lnot ((A \rightarrow B) \land \lnot B)$.

But $P \rightarrow Q$ is also equivalent to $\lnot (P \land \lnot Q)$; so the formula it is simply :

$A \rightarrow ((A \rightarrow B) \rightarrow B)$.

Now, a Natural Deduction proof is quite easy :

(1) $A$ - assumed

(2) $A \rightarrow B$ --- assumed

(3) $B$ --- from (1) and (2) by $\rightarrow$-elimination

(4) $(A \rightarrow B) \rightarrow B$ --- from (2) and (3) by $\rightarrow$-introduction

(5) $A \rightarrow ((A \rightarrow B) \rightarrow B)$ --- from (1) and (4) by $\rightarrow$-introduction.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.