Conditional Proposition Equivalence

In summary: The tests of the implications and use of truth tables do not apply to individual values of ##n## but to the... implications, in this case three of them. ##n## is either < 7, > 7 or = 7.In summary, the equivalence (p → r) ∨(q → r) ≡ (p ∧q) → r can be justified by considering the three implications involved and their respective truth values when p, q, and r are assigned values. It can also be understood by considering specific examples, such as the one provided where p is "n ≥ 7
  • #1
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Hello,

I am confused with the equivalence: (p r) (q r) (p q) r. I checked that truth tables supports it but I cannot imagine an example which justifies it.

Suppose: p says “It is raining”, q says “It is snowing” and r says: “we will close”. So (p r) (q r) becomes “if it is raining then we will close or if it snows then we will close”. Is this not same as saying If it rains or snows or both then we will close? Then why (p q) r and not (p q) r ?

Thanks in advance.

* Also, can you pls. provide an example for p → q ≡ ¬p ∨ q ?
 
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  • #2
Your example for (p r) (q r) would be "either we close when it rains or we close when it snows (or both). "
So, the logical equivalent to that is: if it is both snowing and raining, then we will close.
It is not the same as: if it is snowing or raining, then we close...since this case allows us to close only when it snows and the original statement allows us only to close when it rains.

p → q ≡ ¬p ∨ q is easier if you are comparing when the statements are false.
p → q means if p then q, which is false only when p is true and q is false.
So p → q ≡¬(p ∧ ¬q) which is the same as ¬p ∨ q.
 
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  • #3
I am confused, too. Let's say p and q is never true. Then p might still imply r, but p and q does not.

E.g. I drink a coke (r) if you roll a dice and the number is even (p). I also drink a coke (r) if you roll a number which can be divided by 3 (q).
So [ ( p → r ) ∨ ( q → r ) ] → [ (p ∧ q) → r], i.e. I'll get a coke (r) on 6 (p and q). However, the other direction is false.
If I get a coke (r) on 6 (p and q), one must not conclude that I'll get one on even numbers (p) or multiples of three (q). It could well be, that I only get it on 6.

p → q ≡ ¬p ∨ q might be clearer if we write it with Venn diagrams. Let X denote truth and p → q ≅ Q ⊆ P.
Thus ¬p ∨ q ≅ X \ P ∪ Q ⊆ X \ Q ∪ Q = X ≅ 1, i.e. ( p → q ) → ( ¬p ∨ q ).
For the other direction let ( ¬p ∨ q ) ∧ p = ( ( ¬p ∧ p ) ∨ ( q ∧ p ) ) = 0 ∨ ( q ∧ p ) = ( q ∧ p ) → q.

As an example let's go back to the dice. This time I drink a coke (q) if you roll an even number (p).
This means if I drink a coke (q) you might have rolled an odd number (¬p) or I drink it anyway (q).
 
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  • #4
If it's raining then it's cloudy, so either it's not raining or it's cloudy.
$$p~{\rightarrow}~q~{\equiv}~{\neg} p~{\vee}~q$$

Suppose it's sunny, then it's not raining, and so either it's not sunny or it's not raining
$${\neg} q~{\rightarrow}~{\neg} p~{\equiv}~q~{\vee}~{\neg}p$$
 
  • #5
SamitC said:
* Also, can you pls. provide an example for p → q ≡ ¬p ∨ q ?
In common language, the truth or falsity of ##p \rightarrow q ## is controversial in the case when ##p## is false. For example, Suppose Bob is not 10 feet tall. Is it true that "if Bob is ten feet tall then Bob likes spaghetti" ? You can debate this at a dinner table with friends and hear a variety of opinions.

However, in mathematics, we deal with proof and counterexamples. We also deal with statements that have quantifiers like "for each". A statement asserting something is true "for each..." can be disproven by giving just one counterexample where the statement is false.

Consider the statement "For each real number ##r##, if ##r## is an integer then ##2r## is an integer". In mathematics, we don't want someone to disprove this claim, by using the counterexample ##r = 1/4##.

Conceivably, you could disallow ##r = 1/4## as a counterexample by adding vague language to mathematics about how a counterexample must "apply" to the claim being disproven. However, the elegant way to handle the situation is simply to stipulate that the statement "if ##r## is an integer then ##2r## is an integer" is true when someone gives ##r## a value that is not an integer. This prevents people from disproving the "if...then..." statement by presenting cases where the "if..." part isn't true.

If you understand why it is best to stipulate that "##p \rightarrow q## is true when ##p## is false then I think you will be able to understand the equivalence ##p \rightarrow q \equiv (\lnot p \lor q)##.
 
  • #6
SamitC said:
Hello,

I am confused with the equivalence: (p r) (q r) (p q) r. I checked that truth tables supports it but I cannot imagine an example which justifies it.
?

I don't see the equivalence. Try

p is ##n \ge 7##

q is ##n \le 7##

r is ##n = 7##
 
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  • #7
PeroK said:
I don't see the equivalence. Try

p is ##n \ge 7##

q is ##n \le 7##

r is ##n = 7##
I can see it if I write it as (~r → ~p) ∨ (~r → ~q). If n ≠ 7 then n is not ≤ 7 or (meaning and/or) n is not ≥ 7.
 
  • #8
EnumaElish said:
I can see it if I write it as (~r → ~p) ∨ (~r → ~q). If n ≠ 7 then n is not ≤ 7 or (meaning and/or) n is not ≥ 7.
The tests of the implications and use of truth tables do not apply to individual values of ##n## but to the statements about ##n##.
 
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  • #9
PeroK said:
I don't see the equivalence. Try

p is ##n \ge 7##

q is ##n \le 7##

r is ##n = 7##
In this example, both the implications are not logically sound.
p→ r : false in general
q→r : false in general
(p→ r) ∨ (q→ r) : false in general
(p q) r : true in general.

However, if you force n to be fixed. Then you have two cases:

Case 1- n = 7, r is true, then:
p→ r : true
q→r : true
(p→ r) ∨ (q→ r) : true
(p q) : true
(p q) r : true

Case 2 - n≠7, r is false, then:
Without loss of generality, assume n>7, making p true and q false.
p→ r : false
q→r : true
(p→ r) ∨ (q→ r) : true
(p q) : false
(p q) r : true

It seems like this sort of logic deals in specific cases, where your outcomes are fixed.

PeroK said:
The tests of the implications and use of truth tables do not apply to individual values of n but to the statements about n.
I tend to agree, but it is clear that this problem assumes that r is fixed - either true or false. And the truth status of the other arguments can be known.
 
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  • #10
PeroK said:
The tests of the implications and use of truth tables do not apply to individual values of ##n## but to the statements about ##n##.
Agreed. My comment was based on the propositions you provided.
 
  • #11
PeroK said:
I don't see the equivalence. Try

p is ##n \ge 7##

q is ##n \le 7##

r is ##n = 7##

That is not a counterexample, because expressions like "##n \ge 7##" are not statements. They don't have a definite truth value.

In the context of both common speech and mathematics, a sentence involving a variable such as ##n## is often interpreted as implicitly meaning "for each n". So, to attempt a counterexample, you can try letting ##p##, ##q##, and ##r## each be quantified expressions. (E.g. you could let ##p## denote "for each ##n, n \ge 7##" .)

I think your approach interprets the notation ##p \rightarrow q## as a statement (i.e. as a quantified expression), but it does not interpret the symbol ##p## standing alone as an statement.

You interpret:
##(p \rightarrow r ) ## to mean "For each ##n, n \ge 7 \rightarrow n = 7##".
##(q \rightarrow r) ## to mean "For each ##n, (n \le 7 ) \rightarrow n = 7##".

However this interpretation does not specify how to interpret the symbol "##p##" by itself as a statement. It uses ##p## to denote an expression.

The equivalence ##(p \rightarrow r) \lor (q \rightarrow r) \equiv (p \land q) \rightarrow r## is a claim about symbols ##p##, ##q## and ##r## that denote statements.
 
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  • #12
Stephen Tashi said:
The equivalence ##(p \rightarrow r) \lor (q \rightarrow r) \equiv (p \land q) \rightarrow r## is a claim about symbols ##p##, ##q## and ##r## that denote statements.
That explains it. Thanks.
 

What is "Conditional Proposition Equivalence"?

Conditional Proposition Equivalence is a logical concept that refers to the relationship between two conditional statements. It states that two conditional statements are equivalent if and only if their corresponding antecedents and consequents are logically equivalent.

How is "Conditional Proposition Equivalence" represented?

Conditional Proposition Equivalence is typically represented using the symbol "↔" or the phrase "if and only if". For example, the statement "If it is raining, then the ground is wet" is equivalent to "The ground is wet if and only if it is raining".

What is the difference between "Conditional Proposition Equivalence" and "Material Implication"?

Conditional Proposition Equivalence and Material Implication are both logical concepts that involve conditional statements. However, Material Implication is a weaker form of equivalence, as it only requires the consequent to be true whenever the antecedent is true. Conditional Proposition Equivalence, on the other hand, requires the antecedent and consequent to be logically equivalent.

How is "Conditional Proposition Equivalence" used in logic and mathematics?

Conditional Proposition Equivalence is a fundamental concept in logic and mathematics, and it is often used to prove the validity of logical arguments. It is also used in algebraic manipulations and proof techniques, such as proof by contraposition and proof by contradiction.

What are some examples of "Conditional Proposition Equivalence"?

One example of Conditional Proposition Equivalence is the statement "If it is raining, then the ground is wet". This is equivalent to the statement "The ground is wet if and only if it is raining". Another example is the statement "If John is taller than Mary, then Mary is shorter than John", which is equivalent to "Mary is shorter than John if and only if John is taller than Mary".

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