Is Logical Equivalence of Conditional Statements a valid title for this content?

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Homework Help Overview

The discussion revolves around the logical equivalence of conditional statements, specifically examining the expression (p → q) ∨ (p → r) and its equivalence to p → (q ∨ r).

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the transformation of the logical expressions using laws of logic, with one participant expressing uncertainty about the correctness of their steps. Others suggest alternative methods, such as using a truth table to verify the equivalence.

Discussion Status

Some participants confirm the correctness of the steps taken in the transformation, while others question the necessity of the approach used. There is an exploration of different methods to validate the equivalence, including truth tables and logical laws.

Contextual Notes

There is uncertainty regarding the requirements set by a textbook, which may affect how participants approach the problem.

bonfire09
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Homework Statement


(b) Show that (p → q) ∨ (p→ r) is equivalent to p → (q ∨ r).


Homework Equations




the ~ means negate

The Attempt at a Solution



Im not sure if i did this correctly
(p → q) ∨ (P → r)
(~p∨q) ∨ (~p∨r) used the conditional law p→q equivalent to ~p∨q
((~p∨q)∨~p)∨((~p∨q)∨r)) distributive law
(~p∨q)∨(~p∨q)∨r
(~p∨q)∨r
~p∨(r∨q) associative law
p →(q∨r)
 
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hi bonfire09! :smile:

all your steps are correct

however, after …
bonfire09 said:
(p → q) ∨ (P → r)
(~p∨q) ∨ (~p∨r)

… don't you notice that they're all ∨ ,

so you can rearrange them (using the …?… law), and then use ~p∨~p = ~p :wink:
 
Are you required to do it that way? Setting up an 8 case "truth table" shows that both statements are false in case p= T, q= r= F, and true in all other cases.
 
That I am not sure upon. In Velleman's book he is not so clear about what he wants us to show. But I think what I did suffices.
 

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