Proof of p^(qvr) <=> (p^q)v(p^r)

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The discussion focuses on proving the equivalence of the logical expression p^(qvr) and (p^q)v(p^r) in propositional calculus. Participants suggest using truth tables for a semantic proof, while also exploring alternative methods. There is uncertainty about the definition of semantic versus syntactic proofs, with one participant considering a contradiction approach for the syntactic proof. The conversation encourages sharing ideas to refine understanding and build confidence in the proof process. Overall, the thread highlights the importance of clarity in logical proofs and collaborative problem-solving.
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How do we prove in propositional calculus :

...p^(qvr) <===> (p^q)v(p^r) semantically and syntactically
 
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Well, what have you already come up with?
 
Comparing truth tables will do it quickly and neatly. Are you not allowed to use that method?
 
I had to Google as well, as far as I http://www.rci.rutgers.edu/~cfs/472_html/Logic_KR/proplogic_proofs472.html , using truth tables would be the semantic proof.
 
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Semantical proof without using true tables ,i have one in mind but i am not very positive about it.Then syntactically how about a contradiction you think it could work ,although it looks a bit messy
 
It is not quite clear to me what you mean by a semantical proof, and a syntactical one.
Also, if you would post your idea we can have a look at it. Maybe you are on the right track but just need a last push, or maybe you even got it right but lack the confidence :wink:
 

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...
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