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.
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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.
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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.
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**Final note:** These are logical formula derivations involving quantifiers and implications, showing how premises lead to conclusions in predicate logic.