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I'm sorry for put an image insted of typing it...

infact this is an 2012-exam on Logic. i found the solution of this quiz that wrote by one TA. he wrote just the second line is not valid logically in First Order Logic, but i think this answer is false and the just forth line is the correct one. anyone could judge me and help me? $$\begin{align*}\rm\tag{1}\forall x\;\neg A(x) \to \exists x\;\big(A(x)\to\bot\big) \\\rm\tag{2} \forall x\;\big(A(x)\leftrightarrow B(x)\big)\leftrightarrow\big(\neg\forall x\;A(x)\leftrightarrow\neg\forall x\; B(x)\big) \\\rm\tag{3} \exists x\;\neg\big(A(x)\vee B(x)\big)\to\big(\exists x\;A(x)\vee \exists x\; B(x)\big) \\\rm\tag{4} \exists x\;\big(A(x)\to B(x)\big)\to\big(\exists x\;A(x)\vee \exists x\; B(x)\big) \end{align*}$$

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  • $\begingroup$ For (2), take $A(x)$ to mean $x = 1$ and $B(x)$ to mean $x = 2$, then $A(x)$ and $B(x)$ are not equivalent (lhs of (2) is false), but they are both falsifiable (rhs of (2) is true). $\endgroup$ – Rob Arthan Feb 10 '15 at 0:23
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\begin{align*}\rm\tag{1}\forall x\;\neg A(x) \to \exists x\;\big(A(x)\to\bot\big) \\\forall x \neg A(x)\to \exists\neg A(x)\tag{$\checkmark$} \\[3ex]\rm\tag{2} \forall x\;\big(A(x)\leftrightarrow B(x)\big)\leftrightarrow\big(\neg\forall x\;A(x)\leftrightarrow\neg\forall x\; B(x)\big) \\ \text{see below} \\[2ex]\rm\tag{3} \exists x\;\neg\big(A(x)\vee B(x)\big)\to\big(\exists x\;A(x)\vee \exists x\; B(x)\big) \\\neg \forall x\;\big(A(x)\vee B(x)\big)\to \exists x\;\big(A(x)\vee B(x)\big) \tag{$\color{green}{\checkmark}$} \\[2ex]\rm\tag{4} \exists x\;\big(A(x)\to B(x)\big)\to\big(\exists x\;A(x)\vee \exists x\; B(x)\big) \\ \exists x\;\big(\neg A(x)\vee B(x)\big)\to \big(\exists x\;\neg A(x) \vee \exists x\; B(x)\big) \tag{$\color{red}{\times}$} \end{align*}

The second one takes a little more work to show that you could not get to the consequent from the antecedent.

$\newcommand{\fro}{\leftarrow}\newcommand{\tofro}{\leftrightarrow} \begin{align} \forall x (A\tofro B) & \iff \forall x((A \wedge B)\vee (\neg A\wedge \neg B)) \\ & \implies (\exists x(A) \wedge \exists x(B))\vee(\exists x(\neg A) \wedge \exists x(\neg B)) \\ & \implies (\neg \forall x(\neg A) \wedge \neg\forall x(\neg B))\vee(\neg\forall x( A) \wedge \neg\forall x(B)) \\ \\ \neg\forall x \,A \tofro \neg\forall x\, B & \iff (\neg\forall x\, A\wedge \neg \forall x\, B )\vee(\forall x\, A\wedge \forall x\, B) \end{align}$

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  • $\begingroup$ you means LHS of 2 is false or has typo ? $\endgroup$ – Ali Movagher Feb 10 '15 at 4:18

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