# Why "the conditional in which the antecedent is false" is always true?

• B
Summary:
Gp→Gl is true since the conditional in which the antecedent is false is always true. (Agnishom does not play the guitar).
I'm just learning some basic predicate logic. I found this.

• UD: People
• Gx: x can play the guitar
• l: Lemmy
In the expression ∃xGx→Gl, the scope of the quantifier ∃ is the expression Gx. This translates to If there is a guitarist, Lemmy is a guitarist.

Now this is changed to:

∃x (Gx→Gl), we have changed the scope of the quantifier to the entire expression. The sentence now means, There is a person x such that if x is a guitarist, Lemmy is a guitarist.

I think I follow the above, but the next part:

You might notice that this sentence is true because non-Guitarists exist.

For example, let p be Agnishom. Gp→Gl is true since the conditional in which the antecedent is false is always true. (Agnishom does not play the guitar). Since someone, namely p, satisfies the sentence, ∃x(Gx→Gl) is true.

Gp is FALSE (When p = Agnishom, I think)

I expected, only if Gx is true then (Gx→Gl) is true.

Why is (Gp→Gl) TRUE when Gp is FALSE?? (Agnishom does not play the guitar)

Thanks for any tips or pointers. I want to understand this before I continue. It seems important.

jedishrfu

Stephen Tashi
Consider the proposition:

For each number x, if x > 3 then x > 2.

Would you want x = 1 to be an example showing that the above proposition is False?

If not, then we must use the convention that "if 1 > 3 then 1 > 2" is True.

jedishrfu and FactChecker

Sorry, but I don't get it! Could you elaborate?

p is not a guitarist, by definition
Gp is FALSE (When p = Agnishom, I think)

I read Gx→Gl : if Gx is true, then Gl is true ( and the obverse is : when Gx is false, then Gl is false)

1>3 is false
1>2 is false

but I don't follow the relationship to the guitarists!

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Stephen Tashi
I read Gx→Gl : if Gx is true, then Gl is true ( and the obverse is : when Gx is false, then Gl is false)
Where do you read the obverse? (I can't use the link you gave because it requires a login.)

I assume the obverse. Is that not so?

UD : all people

Some of them are guitarists, but I have no way of knowing that Lemmy is or is not able to play.

If I want to know if Lemmy is a guitarist, I will look up his name in a list of guitar players.

Python:
g_players = ['Jane', 'Jenny', ...... , 'Paul', 'Percival']

def is_a_guitarist(name):
if name in g_players:
print(name, 'is a guitarist')
else:
print(name, 'is not a guitarist')

Why assume someone is a guitarist because someone else is?

Likewise, if I want to know if a number is greater than 3:

Python:
if num > 3:
print(num, 'is greater than 3')

Why would I make counter-intuitive assumptions?

Stephen Tashi
I assume the obverse. Is that not so?

If you are asking whether the Truth of ##A \implies B## guarantees the Truth of ##\lnot A \implies \lnot B##, it does not. For example:

For each whole number x, if x is divisible by 14 then x is divisible by 7.

Consider the case of x = 21.

The Truth table for ##A \implies B## is different than the state transistion for a computer instruction. The usual convention in computer languages is that the instruction "If A then do B" does not execute instruction B if condition A is False. However, this does not prohibit a different instruction from executing instruction B in the case when condition A is False.

FactChecker
Gold Member
I read Gx→Gl : if Gx is true, then Gl is true ( and the obverse is : when Gx is false, then Gl is false)
I believe that this is not the valid obverse of ##Gx\rightarrow Gl##

stevendaryl
Staff Emeritus
This is maybe a way to think of it.

Let’s make up an ordering on truth values. Say that true is greater than false. There are only two truth values, so that tells us everything we need to know about the ordering. Then for any two sentences ##A## and ##B##, we can say that ##A \leq B## if the truth value of ##A## is less than or equal to the truth value of ##B##. Either they are both true, or both false, or ##A## is false while ##B## is true.

Now, we can establish some facts about ##\leq##.

1. If we know ##A \leq B## and we know ##A## is true, then we know ##B## is true.
2. If we know that ##A## is false, then we know that ##A \leq B##, regardless of what ##B## is.
3. If we know that ##B## is true, then we know ##A \leq B##, regardless of what ##A## is.

If this makes sense to you, then you can just interpret ##A\rightarrow B## as ##A \leq B##.

WWGD and Office_Shredder
If this makes sense to you, then you can just interpret ##A\rightarrow B## as ##A \leq B##.
Some other expressions:
##A \leq B##
##A \geq B##
##A<B##
##A>B##
##A=B##
##A \neq B##

stevendaryl
Staff Emeritus
Some other expressions:
##A \leq B##
##A \geq B##
##A<B##
##A>B##
##A=B##
##A \neq B##
In terms of implication, those would be written as

##A \rightarrow B##
##B\rightarrow A##
##(A \rightarrow B ) \wedge \neg (A \leftrightarrow B)##
##(B \rightarrow A ) \wedge \neg (A \leftrightarrow B)##
##A\leftrightarrow B##
##\neg (A\leftrightarrow B)##

WWGD
Gold Member
Some other expressions:
##A \leq B##
##A \geq B##
##A<B##
##A>B##
##A=B##
##A \neq B##
I wonder if it is a stretch to argue that implication "rewards" non-decreasing truth values and punishes decreasing ones, i.e., we don't "reward" ( by assigning a truth value T) a sequence where we start with a true statement and derive a false one

Not sure whether you meant so say something precise or not. At any rate, an expression such as:
##A \leq B \leq C##
came to my mind. Upon writing truth table (if I didn't happen to make a mistake), it doesn't seem to be equivalent to either of:
##(A \rightarrow B) \rightarrow C##
##A \rightarrow (B \rightarrow C)##

[of course they are equivalent to ##(A \leq B) \leq C## and ##A \leq (B \leq C)## respectively]

Note:

Though probably one would note though that if implication did happen to be associative, then ##A \leq B \leq C## would not be particularly good notation. Since ##A \wedge B \wedge C## is written as a short-hand for ##(A \wedge B) \wedge C## or ##A \wedge (B \wedge C)##. And similarly for ##A \vee B \vee C## too.

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pbuk
Gold Member
I wonder if it is a stretch to argue that implication "rewards" non-decreasing truth values and punishes decreasing ones, i.e., we don't "reward" ( by assigning a truth value T) a sequence where we start with a true statement and derive a false one
Yes that is too much of a stretch. Predicate logic is a formal system in which things mean exactly what we define them to mean, no more and no less. We define the relationship ## A \implies B ## as corresponding to the following truth table:
Code:
A | B | A ⇒ B
-------------
F   F |   T
F   T |   T
T   F |   F
T   T |   T

Trying to rationalize this with the baggage that comes with the word "implies" from everyday speech is a waste of time. Predicate logic is not a tool for encoding everyday speech, it is a tool for helping us construct formal statements whose truth can be tested.

WWGD
Gold Member
Yes that is too much of a stretch. Predicate logic is a formal system in which things mean exactly what we define them to mean, no more and no less. We define the relationship ## A \implies B ## as corresponding to the following truth table:
Code:
A | B | A ⇒ B
-------------
F   F |   T
F   T |   T
T   F |   F
T   T |   T

Trying to rationalize this with the baggage that comes with the word "implies" from everyday speech is a waste of time. Predicate logic is not a tool for encoding everyday speech, it is a tool for helping us construct formal statements whose truth can be tested.
How does that truth table output describe/model implication better than any of the other 15 choices then ?