MHB How to Prove Predicate Logic Validity with Induction?

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To prove the validity of the given predicate logic argument using induction, one must first clarify the meaning of the symbols, particularly the "!" operator. The discussion emphasizes the importance of showing progress in problem-solving to facilitate effective assistance from others. Participants are encouraged to share their attempts and reasoning to avoid redundant suggestions. The thread highlights the need for a structured approach to induction in logic proofs. Overall, the focus is on collaborative learning and clarity in communication to tackle the problem effectively.
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How would I go about proving that the argument below is valid using the induction method?
(∃x)[P(x)!Q(x)]^(∀y)[Q(y)!R(y)]^(∀x)P(x)!(∃x)R(x)

Thank you very much in advance!
 
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Hello and welcome to MHB, Voehet! :D

We ask that our users show their progress when posting questions, and that way our helpers can see where you are stuck or may be going astray and will be able to post the best help possible without potentially making a suggestion which you have already tried, which would waste your time and that of the helper.

Can you post what you have done so far?
 
Voehet said:
How would I go about proving that the argument below is valid using the induction method?
(∃x)[P(x)!Q(x)]^(∀y)[Q(y)!R(y)]^(∀x)P(x)!(∃x)R(x)
What does ! stand for?
 

Thread 'Formal derivation of statement from Peano Arithmetic system'

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...
View full post »

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