MHB Conditional proof for multiple quantifier

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To prove ((Ǝx) F(x) → (Ǝx) G(x)) using conditional proof, start with the premises: ((Ǝx) F(x) → (∀z) H(z)) and H(a) → G(b). The discussion suggests that a formal proof in logical calculus may be necessary for clarity. A reference link is provided for additional guidance on formal proofs. Understanding the relationship between existential and universal quantifiers is crucial in this context.
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Hi, I don't know how to prove ((Ǝx) F(x) →(Ǝx) (G(x)) with conditional proof from:
((Ǝx) F(x) → (∀z) H(z))
H(a) →G(b)

Thanks
 
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What kind of proof do you have in mind? If you are talking about a formal proof in some logical calculus, then please see https://driven2services.com/staging/mh/index.php?threads/29/.
 
I'm taking a look at intuitionistic propositional logic (IPL). Basically it exclude Double Negation Elimination (DNE) from the set of axiom schemas replacing it with Ex falso quodlibet: ⊥ → p for any proposition p (including both atomic and composite propositions). In IPL, for instance, the Law of Excluded Middle (LEM) p ∨ ¬p is no longer a theorem. My question: aside from the logic formal perspective, is IPL supposed to model/address some specific "kind of world" ? Thanks.
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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