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The Bernays–Schönfinkel class is not the propositional fragment of first-order logic because the sentences in it are not propositional! They involve quantifiers and variables. Now it's true that a sentence $\varphi$ in the Bernays–Schönfinkel class can be transformed into a propositional sentence $\widehat{\varphi}$: (1) introduce constant symbols as ...


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You made a good start but you've missed a $\lnot$ before $b$ in the last formula (EDIT: now fixed). After fixing that, use the distributivity law $(x \land y) \lor z = (x \lor z) \land (y \lor z)$ to lift the conjunction up to the top: \begin{align*} (a \implies b) \implies (c \implies d) &= \lnot(\lnot a \lor b) \lor (\lnot c \lor d) \\ &= (a \...


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Firstly, let us make clear that there are many different but equivalent sets of axiom schemata for propositional calculus, which serve the same purpose: from them, we can prove the same theorems. Some of them include only the least possible number of axiom schemata. Others have redundant axiom schemata, that can be proved from the rest and thus, are not ...


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Since $T$ is complete, for each propositional variable $P$, exactly one of $P$ and $\neg P$ is in $T$. So this gives you a truth assignment: let $V$ be the truth assignment that says a propositional variable is true if it is in $T$ and false if its negation is in $T$. Now you can use the fact that $T$ is closed under tautological consequence to show that ...


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Hint The two conditions are equivalent. For one side, if $\alpha$ is a tautology, that means that $\alpha$ is always true. Thus, the cases when $\{ γ,α \}$ is satisfied are exactly the cases when $\{ γ \}$ is. Similar for the other part.


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All I can think of is to assume $B$, and then assume $A\supset C$, and somehow maybe get to ${\sim}(A\supset C)$. Well, as you have found, you do not need to assume $A\supset C$ when you have derived it. Also that "somehow maybe get" is "modus ponens" or "conditional elimination", by which $\neg(A\rightarrow C)$ is inferred ...


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