Consider the set of inference rules for first order logic (analogous to the ones listed here : http://en.wikipedia.org/wiki/Sequent_calculus#Inference_rules)

I am stuck in proving the following rule

$$\vdash_{\gamma} \neg \forall x.\phi \implies \exists x. \neg \phi $$

I think it is easy to do this using the notion of soundness and completeness and checking that the left formula is valid when the right is.

However I am not able to prove it using just the formalism of manipulating proof trees with inference rules. Somehow I do not see how to get rid of the negation in $\neg \forall x.\phi$ without applying the rule I want to prove.

Any hints?


Would the following work?

1 $\vdash \neg \forall x. \phi$ | Hypothesis

2 $\vdash \neg \phi \implies \exists x. \neg \phi$ by existential generalization

3 $\vdash \neg \exists x. \neg \phi \implies \phi$ by 1,Contraposition

4 $\neg \exists x. \neg \phi \vdash \phi$ by 3

5 $\neg \exists x. \neg \phi \vdash \forall x. \phi$ by 4,Universal Generalization

6 $\vdash \neg \exists x. \neg \phi \implies \forall x. \phi$ by 5,Deduction

7 $\vdash \neg \forall x. \phi \implies \exists x. \neg \phi$ by 6,Contraposition


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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