Logic Proofs 260607

1. **Problem 1:** Prove the implication $$((\forall x Fx \wedge \forall x(Fx\to Gx))\to \forall x Gx)$$. 2. **Formula used:** Universal quantifier rules and implication elimination. 3. **Step-by-step:** 1. Assume $$\forall x Fx \wedge \forall x(Fx\to Gx)$$. 2. From the conjunction, extract $$\forall x Fx$$ and $$\forall x(Fx\to Gx)$$. 3. By universal instantiation, for an arbitrary element $$x$$, we have $$Fx$$ and $$Fx\to Gx$$. 4. By modus ponens, from $$Fx$$ and $$Fx\to Gx$$, conclude $$Gx$$. 5. Since $$x$$ was arbitrary, generalize to $$\forall x Gx$$. 6. Therefore, $$((\forall x Fx \wedge \forall x(Fx\to Gx))\to \forall x Gx)$$ holds. --- 1. **Problem 2:** Analyze $$\forall x((Fx\to Gx)\lor (Gx\to Fx))$$ and related implications. 2. **Formula used:** Logical disjunction, implication, negation, and universal quantification. 3. **Step-by-step:** 1. Assume $$Fx$$. 2. From the disjunction $$((Gx\to Fx)\lor (Fx\to Gx))$$, consider cases. 3. If $$Gx$$, then $$Fx\to Gx$$ holds. 4. If $$\neg Gx$$, then $$Gx\to Fx$$ holds vacuously. 5. The statement $$((Gx\to Fx)\lor (Fx\to Gx))\to \neg ((Gx\to Fx)\wedge (Fx\to Gx))$$ shows that both implications cannot hold simultaneously. 6. Hence, $$\forall x((Fx\to Gx)\lor (Gx\to Fx))$$ is consistent with the negation of their conjunction. --- 1. **Problem 3:** Prove $$((\forall x Fx \wedge \forall x \neg Gx)\to \neg \forall x(Fx\to Gx))$$. 2. **Formula used:** Universal quantifiers, negation, and implication. 3. **Step-by-step:** 1. Assume $$\forall x Fx \wedge \forall x \neg Gx$$. 2. From the conjunction, extract $$\forall x Fx$$ and $$\forall x \neg Gx$$. 3. By universal instantiation, for an arbitrary $$x$$, $$Fx$$ and $$\neg Gx$$ hold. 4. Since $$Fx$$ is true and $$Gx$$ is false, $$Fx\to Gx$$ is false for that $$x$$. 5. Therefore, $$\forall x(Fx\to Gx)$$ is false, so $$\neg \forall x(Fx\to Gx)$$ holds. 6. Hence, $$((\forall x Fx \wedge \forall x \neg Gx)\to \neg \forall x(Fx\to Gx))$$ is proven.