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1d
answered $p \to (q\vee\neg r), \neg q, r ⊢ \neg p$ - Natural deduction- elimination with $\neg$ operator
1d
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.
Jan
19
comment Associativity and De Morgan's for more than 2 literals
No, it's logically significant, because if (A∧B∧C) is an expression, than the rule of uniform substitution fails. The rule of uniform substitution is important for having an axiomatic system of propositional logic. If say A∧B is meaningful (which you implied), and we have the rule of uniform substitution, then substituting B with CvD we obtain A∧CvD. Suppose A=0, C=0, D=1. Then [A∧(CvD)]=0, while [(A∧C)vD]=1. So, I started with an expression assumed as meaningful and deduced a contradiction, which renders either the expression or the rule of uniform substitution as not meaningful.
Jan
17
comment Associativity and De Morgan's for more than 2 literals
No, take a look at the formation rules. Also, the rule of uniform substitution would fail if (A∧B∧C) were meaningful.
Jan
12
answered Is there a law that you can add or multiply to both sides of an equation?
Jan
3
comment Is the Cardinality of the Set of Contingent Propositions the Same as the Cardinality of the Set of Tautologies?
The constructions require variables and connectives. But that said, I don't think that's a problem. I don't see how you infer from (p∨¬p) as a tautology to that inequality. Also, if ϕ is a tautology, then (p∨ϕ) is also a tautology.
Dec
30
comment Puzzle: Give an algorithm for finding a frog that jumps along the number line
Leave the frogs alone. Experiment on yourselves.
Dec
23
comment Formula to map 1 to 5… to 5 to 1
"But there is an infinite number of answers !" And an infinity of them are probably useless for the vast majority of applications that anyone would want and not very interesting to the vast majority of mathematicians!
Dec
23
comment Is this a correct solution to determining which of two people is the liar using one question?
Suppose that the liar says that he is a liar. Then he has told the truth about himself and thus not lied. But then he is not a liar, and thus his statement about himself is false, and consequently he is a liar.
Dec
23
revised Prove $( \lnot C \implies \lnot B) \implies (B \implies C)$ without the Deduction Theorem
added 372 characters in body
Dec
23
comment Can an electronic computer (stochastic) simulation be used as a “formal” proof of a tedious mathematical problem?
Who are the peers in question here?