# Does anyone know how to prove ~ ∀x (Ax→Bx) from Ǝx(Ax & ~Bx)?

1. Ǝx(Ax & ~Bx) Premise
2. SHOW: ~ ∀x (Ax→Bx)

I really appreciate anyone who could help

The instructions for the homework were to Prove that the obverse of a particular negative entails the particular negative itself. i.e. that ‘Some A are not B’ entails ‘Not all A are B’. Then use morgans law. The book I am using is Critical Thinking by Moor-and Parker 11th edition.

• P & ~Q is equivalen to ~ (P→Q). Dec 22, 2018 at 11:43
• You may want to include what rules you are permitted to use and the text you are using. Welcome to Philosophy. Dec 22, 2018 at 13:20

One way to approach this is to add an assumption with the goal of eliminating the existential quantifier by giving the variable x in the premise a name, say a.

Then add a second assumption of Aa→Ba with the goal to arrive at a contradiction. You should be able to easily reach that contradiction with the previous assumption. That will allow you to discharge the second assumption by introducing a negation.

With that negative conditional, introduce an existential quantifier on the name a calling the variable x. Now that you no longer have the name a, you can complete the elimination of the existential quantifier from the premise and discharge the first assumption.

A change of quantifier is all that is needed at this point to complete the proof.

I was able to complete the proof in 11 lines using Klement's Fitch-style natural deduction proof checker using the following rules: conjunction elimination, conditional elimination, contradiction introduction, negation introduction, existential elimination and introduction and change of quantifiers.

What you are using may require using different rules. The forallx text linked below may also provide a supplemental reference to whatever text you are using.

The following rule is available as Rule 12 Implication (IMPL) in the Group II Rules in Moore and Parker's Critical Thinking 9th Edition:

This rule allows us to change a conditional into a disjunction and vice versa. (P → Q) ↔ (~P ∨ Q)

Using this we could also start with the premise, Make an assumption to attempt to eliminate the existential quantifier over the variable x to obtain Aa & ~Ba using the specific name a. Use DeMorgan to transform that into ~(~Aa v Ba) Use rule 12 to transform that into ~(Aa → Ba). Introduce the existential quantifier to replace the name a with x. This will allow us to discharge the assumption and then with a change of quantifiers obtain the final result.

This may be another way to obtain the result.

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

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/

• The instructions for the homework were to Prove that the obverse of a particular negative entails the particular negative itself. i.e. that ‘Some A are not B’ entails ‘Not all A are B’. Then use morgans law. The book I am using is Critical Thinking by Moor-and Parker 11th edition.
– rom
Dec 22, 2018 at 21:07
• @rom I added an alternate approach using a rule I found in an older edition of the text you are using. Dec 22, 2018 at 23:42
• Thank you so much for the help, this seems accurate to the homework.
– rom
Dec 23, 2018 at 4:43