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Mar
25
comment Completness and Set of Result of One Set ?!?
@CarlMummert Also, are you sure about that? The normal deductive system just has tautologies in the object language. The set in question here has statements which evaluate to true and statements which evaluate to false.
Mar
25
comment Completness and Set of Result of One Set ?!?
@CarlMummert Well, the normal deductive system isn't complete according to the definition referenced by the author in the comments, because no contingent proposition is provable, nor is it's negation provable. We can pick any propositional variable as $\phi$ there If the variable was provable, then by assigning it a truth value of "false" we could prove a false statement, and classical logic would be unsound. Similarly, if the negation of the variable was provable, then by assigning the variable a truth value of "true" we could prove a false statement, and classical logic would be unsound.
Mar
25
comment Completness and Set of Result of One Set ?!?
@LoveMathContest If you can't prove anything, then the logical results of the set is empty. The empty set is enumerable since it corresponds to the ordinal 0.
Mar
25
revised Completness and Set of Result of One Set ?!?
added 43 characters in body
Mar
25
answered Completness and Set of Result of One Set ?!?
Mar
22
answered Can I use two inferred clauses to get the empty set?
Mar
16
comment Logic question in propositional calculus
That is not a formula in propositional calculus. You can't prove this in any propositional calculus, since "..." is not part of any vocabulary for any propositional calculus, and thus your statement does not follow the formation rules for a well-formed formula.
Feb
24
answered Proving using axioms of propositional logic
Feb
5
answered When proving the Hypothetical Syllogism inference rule, why must you assume that p is true?
Jan
29
answered $p \to (q\vee\neg r), \neg q, r ⊢ \neg p$ - Natural deduction- elimination with $\neg$ operator
Jan
29
comment What is the correct form of De Morgan's Law in logic?
This can't be write. We're on the internet after all. /s
Jan
27
answered Principle of explosion: Other arguments?
Jan
27
comment Principle of explosion: Other arguments?
If you look at classes axiom sets for classical propositional calculus, say in the appendix of A. N. Prior's book Formal Logic, you can find a bunch of them where it is possible to prove the non-organic theorem CNCpNNpq. Since Kpq is defined as NCpNq, this means that such axiomatic systems can argue for CKpNpq, by proving CNCpNNpq.
Jan
27
answered Principle of explosion: Other arguments?
Jan
23
awarded  Popular Question
Jan
23
comment Meaning of symbols $\vdash$ and $ \models$
@MauroALLEGRANZA That's not exactly true. Some people have tried to work out a set theory with relevant logic as a background. There is also fuzzy subset theory, which has infinite-valued logic as a background. Neither have the Deduction Meta-Theorem (in the sense that if $\gamma$, $\alpha$ $\vdash$ $\beta$, then $\gamma$ $\vdash$ C $\alpha$ $\beta$ is a meta-theorem).
Jan
22
answered Meaning of symbols $\vdash$ and $ \models$
Jan
19
comment Associativity and De Morgan's for more than 2 literals
@AndreyPortnoy I'm not sure.
Jan
19
answered Associativity and De Morgan's for more than 2 literals
Jan
19
comment Associativity and De Morgan's for more than 2 literals
You can only use ∧ and ∨ in this way, if we have a system where the only binary expression is either ∧ or ∨ (but you can't have both around). Associativity only ensures that such expressions are meaningful when we have a pure semigroup. If we have two associative operations around, such as the two-valued logical operations "∧" and "∨", then an expression like A∧B∨C can be ambiguous. It's not associativity which makes such work, but rather that for any binary operations B$_1$, B$_2$, for all x, y, z, ((xB$_1$y)B$_2$z)=(xB$_1$(yB$_2$z)). Associativity is a special case of that.