Conditional Proposition Equivalence

[SamitC]

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 ?

 

[RUber]

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.

 

[fresh_42]

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).

 

[xxxx0xxxx]

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
$${\neg} q~{\rightarrow}~{\neg} p~{\equiv}~q~{\vee}~{\neg}p$$

 

[Stephen Tashi]

* 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 $p \rightarrow q \equiv (\lnot p \lor q)$.

 

[PeroK]

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 $n = 7$

 

[EnumaElish]

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.

 

[PeroK]

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 $n$.

 

[RUber]

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.

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.

 

[EnumaElish]

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.

 

[Stephen Tashi]

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.

 

[PeroK]

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.