MHB Simplify Logic using One Connective: Solving ((q→p) ^ (p→r))→(r→q)

  • Thread starter Thread starter c4nn3t
  • Start date Start date
  • Tags Tags
    Logic
AI Thread Summary
The discussion focuses on simplifying the logical expression ((q→p) ^ (p→r))→(r→q) using only one logical connective. The user outlines their workflow, converting the expression into a disjunctive normal form (DNF) but struggles to reduce it to a single connective. Another participant suggests leveraging the identity property of logical conjunction to simplify the expression further. They emphasize that the middle disjunct should be corrected to (p ∧ ¬r). The conversation highlights techniques for logical simplification and the importance of applying logical identities effectively.
c4nn3t
Messages
3
Reaction score
0
So I've got

((q→p) ^ (p→r))→(r→q)

And I have to simplify it down as much as possible using only one logical connective for the end result (not one type, just one total). Here's been my workflow so far:

((¬q v p) ^ (¬p v r))→(¬r v q)
¬((¬q v p) ^ (¬p v r)) v (¬r v q)
¬(¬q v p) v ¬( ¬p v r)) v (¬r v q)
(q ^ ¬p) v (p ^ r) v (¬r v q)

From here, I'm not sure how to boil down this DNF form into a single 'x and/or/implies y'

Mucho thanks in advance ;)
 
Physics news on Phys.org
c4nn3t said:
(q ^ ¬p) v (p ^ r) v (¬r v q)
The middle disjunct should be $p\land\neg r$.

Use the fact that $(x\land y)\lor x=x$. Indeed,
\[
(x\land y)\lor x=(x\land y)\lor (x\land 1)=x\land (y\lor 1)=x\land 1=x.
\]
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
Back
Top