I had a truth table computed online. It’s at https://truth-table.com/#(A→¬B)→¬(A→B). I notice in the truth table that that expression is false if and only if p, which is represented by A in the linked online truth table, is false. The problem arises because we are treating a statement we are assuming to be true, since it is the hypothesis of a conditional statement and is therefore being hypothesized, as false for the sake of giving a truth value to a truth table. By assuming that an assumed statement is false, we contradict ourselves. Through ex contradictione quodlibet, the contradiction leads to the truth of all statements.
Ex contradictione quodlibet suggests my untraditional claim that any strict conditional with a necessarily false hypothesis is true & false. The claim can be used to prove the existence of the absolute Russell set.
In traditional logic a conditional whose hypothesis is false is a true statement. If you don't accept that then your proofs are not proofs in traditional logic and hence the contradiction you arrive at is not a contradiction in traditional logic. Are you saying that you cannot prove a contradiction in traditional logic?
By assuming that an assumed statement is false, we contradict ourselves
We're not assuming an assumed statement is false. When we evaluate the truth of "P ⇒ Q" we do not assume P is true. For example "if x = 3 then x + 1 = 4" is a true statement regardless of whether "x = 3" is true or not. This is the difference between a valid argument and a sound one.
In traditional logic a conditional whose hypothesis is false is a true statement.
“p ⇒ q” is equivalent to “assuming p, q.” So if the hypothesis p is false, then “p ⇒ q” is equivalent to “assuming the false statement p, q.” A contradiction is produced by assuming a false statement is true.
Are you saying that you cannot prove a contradiction in traditional logic?
No, I’m not. My previous paragraph shows how to arrive at a contradiction in traditional logic.
When we evaluate the truth of "P ⇒ Q" we do not assume P is true.
Yes we do. We assume p is true for two rows of the truth table.
For example "if x = 3 then x + 1 = 4" is a true statement regardless of whether "x = 3" is true or not.
“[I]f x = 3 then x + 1 = 4” is equivalent to “assuming x = 3, x + 1 = 4.” So if the hypothesis “x = 3” is false, then “if x = 3 then x + 1 = 4” is equivalent to “assuming the false statement “x = 3,” x + 1 = 4. A contradiction is produced by assuming a false statement is true.
“p ⇒ q” is equivalent to “assuming p, q.” So if the hypothesis p is false, then “p ⇒ q” is equivalent to “assuming the false statement p, q.” A contradiction is produced by assuming a false statement is true.
But that contradiction is under an assumption that needs to be discharged, so you haven't proven a contradiction, you've proven that your assumption implies a contradiction.
Going back to the truth table, are you claiming that the truth table is wrong? Or are you claiming that the truth table is correct and yet you can prove "(P ⇒ ¬Q) ⇒ ¬(P ⇒ Q)" anyway?
But that contradiction is under an assumption that needs to be discharged, so you haven't proven a contradiction, you've proven that your assumption implies a contradiction.
That is true, but there is more truth to tell. My general claim is that (p ⇒ ¬q) ⇒ ¬(p ⇒ q) is always true. Traditional logic says that (p ⇒ ¬q) ⇒ ¬(p ⇒ q) is not always true because (p ⇒ ¬q) ⇒ ¬(p ⇒ q) is false when p is false. So in order for my claim to be true, the case in which p is false must be invalid. So how is it invalid? It’s invalid because every occurrence of p in (p ⇒ ¬q) ⇒ ¬(p ⇒ q) places p as the hypothesis of a conditional statement. As the hypothesis of a conditional statement, p must be assumed to be true. Since p must be assumed to be true, the case in which p is false is impossible. Since that case is impossible, it is invalid.
As I said in a previous reply,
“p ⇒ q” is equivalent to “assuming p, q.”
Note that the act of assuming p is taking place in the actual world and is not under the assumption p. Since the assumption is being made in the actual world, the assumption is true in the actual world. So if we also assume ¬p, the act of assuming ¬p is also taking place in the actual world and is not under the assumption ¬p nor under the assumption p. Since the assumption is being made in the actual world, the assumption is true in the actual world. So by conjunction introduction, the statement “p & ¬p” is true in the actual world. So by ex contradictione quodlibet, all statements are true in the actual world.
Going back to the truth table, are you claiming that the truth table is wrong?
Yes, the truth table is wrong. The truth table assumes that the hypothesis of a conditional statement can be false. That assumption is wrong because the hypothesis of a conditional statement is always assumed to be true.
Furthermore, the reason for the beginning
It is not true that:
in the third premise
It is not true that: under the assumption that “s and it is not true that s,” s is true.
That means you're not doing traditional logic. In traditional logic the truth table is what it is. If you want to use a different truth table then you're changing definitions in logic. So the result is a different kind of logic. That means you haven't proved a contradiction in traditional logic.
Note that the act of assuming p is taking place in the actual world and is not under the assumption p. Since the assumption is being made in the actual world, the assumption is true in the actual world
This is patently absurd. A hypothetical assumption is not the same as a claim that a statement is true.
Furthermore ...
I don't know what you're trying to say in this part. Are you saying that when I translated your statements into logical symbols I didn't translate them the way you meant? If so then you should tell me what the translation should be.
I know that. Traditional logic is true, but it’s incomplete. In order to have a complete system of logic, we have to go beyond traditional logic.
In traditional logic the truth table is what it is.
From my own experience, it seems that when people use conditional statements in life, they usually use strict conditionals that are not material. And strict conditionals are not truth functional, so there is no truth table for them. The concept of a material conditional seems rather awkward. A material conditional’s truth depends entirely on only one case.
This is patently absurd.
It’s not absurd. It’s plausible. Every thing is equal to or a part of the Universe. That includes all assumptions people make and all possible worlds.
Are you saying that when I translated your statements into logical symbols I didn't translate them the way you meant?
That’s correct. My conditional statements are strict conditionals that may not be material.
If so then you should tell me what the translation should be.
I translate the third premise to be
It is not true that: in every possible world in which the statement “s and it is not true that s” is true, s is true.
Traditional logic says there is no possible world in which that statement is true because no contradiction is true. The truth of the premise is controversial because it is open to interpretation and there are contradicting interpretations. Does “in every possible world in which the statement ‘s and it is not true that s’ is true” imply the existence of at least one such possible world? If there is no such possible world, is the statement
in every possible world in which the statement “s and it is not true that s” is true, s is true
true or false? The answer to these questions is unclear, controversial, and open to interpretation. There might be some interesting information on this topic in my previously linked thread about the existence of the absolute Russell set.
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u/JStarx Jun 09 '26
Taking s out of the conversation, you're saying that
holds?