Subjects logic

Predicate Logic 7867B2

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Question: БИЛЕТ 24.5.2024. 1) $(\exists x Fx \wedge \forall x(Ex\to Gx))\to \forall x Gx$ $\forall x Ex \wedge \forall x(Fx\to Gx)$ $\forall x Fx$ $Fx$ $Gx$ $\forall x Gx$ $\forall x(Fx\to Gx) \wedge \forall x(Ex\to Gx))\to \forall x Gx$ 2) $\forall x((Ex\to Gx)\lor (Gx\to Fx))$ $Ex$ $Gx\to Fx$ $(Gx\to Fx)\lor (Fx\to Gx)$ $Fx\to Gx$ $(Gx\to Fx)\wedge (Fx\to Gx)$ $(Gx\to Fx)\wedge (Fx\to Gx)\to \forall x((Gx\to Fx)\wedge (Fx\to Gx))$ 3) $(\forall x Fx \wedge \forall x \neg Gx)\to \neg \exists x(Fx\to Gx)$ $\forall x Fx \wedge \forall x \neg Gx$ $\forall x Fx$ $Fx$ $Gx$ $\neg Gx$ $\neg \exists x(Fx\to Gx)$ $\forall x Fx \wedge \forall x \neg Gx\to \neg \exists x(Fx\to Gx)$
1. **Problem 1:** Prove or analyze the formula $(\exists x Fx \wedge \forall x(Ex\to Gx))\to \forall x Gx$. 2. **Step 1:** Understand the formula components. - $\exists x Fx$ means there exists some $x$ such that $F(x)$ is true. - $\forall x(Ex\to Gx)$ means for all $x$, if $E(x)$ then $G(x)$. - The implication states if both these hold, then for all $x$, $G(x)$ holds. 3. **Step 2:** Given premises: - $\forall x Ex \wedge \forall x(Fx\to Gx)$ - $\forall x Fx$ 4. **Step 3:** From $\forall x Fx$, for any $x$, $F(x)$ is true. 5. **Step 4:** From $\forall x(Fx\to Gx)$ and $F(x)$, by modus ponens, $G(x)$ follows. 6. **Step 5:** Since $x$ was arbitrary, $\forall x Gx$ holds. 7. **Step 6:** Therefore, the implication $(\exists x Fx \wedge \forall x(Ex\to Gx))\to \forall x Gx$ is supported by the premises. --- 1. **Problem 2:** Analyze $\forall x((Ex\to Gx)\lor (Gx\to Fx))$ and related formulas. 2. **Step 1:** Given $Ex$, and $Gx\to Fx$. 3. **Step 2:** Consider the disjunction $(Gx\to Fx)\lor (Fx\to Gx)$. 4. **Step 3:** Also given $Fx\to Gx$. 5. **Step 4:** The conjunction $(Gx\to Fx)\wedge (Fx\to Gx)$ means $Gx$ and $Fx$ are logically equivalent. 6. **Step 5:** The implication $(Gx\to Fx)\wedge (Fx\to Gx)\to \forall x((Gx\to Fx)\wedge (Fx\to Gx))$ states that if the equivalence holds for some $x$, it holds for all. --- 1. **Problem 3:** Analyze $(\forall x Fx \wedge \forall x \neg Gx)\to \neg \exists x(Fx\to Gx)$. 2. **Step 1:** Given $\forall x Fx$ and $\forall x \neg Gx$. 3. **Step 2:** From $\forall x Fx$, $F(x)$ is true for all $x$. 4. **Step 3:** From $\forall x \neg Gx$, $G(x)$ is false for all $x$. 5. **Step 4:** The implication $Fx\to Gx$ is false for all $x$ because $F(x)$ is true but $G(x)$ is false. 6. **Step 5:** Therefore, $\exists x(Fx\to Gx)$ is false, so $\neg \exists x(Fx\to Gx)$ is true. 7. **Step 6:** Hence, $(\forall x Fx \wedge \forall x \neg Gx)\to \neg \exists x(Fx\to Gx)$ holds. --- **Final note:** These are logical formula derivations involving quantifiers and implications, showing how premises lead to conclusions in predicate logic.
$(\exists x Fx \wedge \forall x(Ex\to Gx))\to \forall x Gx$$\forall x Ex \wedge \forall x(Fx\to Gx)$$\forall x((Ex\to Gx)\lor (Gx\to Fx))$$(Gx\to Fx)\wedge (Fx\to Gx)$$(\forall x Fx \wedge \forall x \neg Gx)\to \neg \exists x(Fx\to Gx)$