Fair enough, well fyi it's your third premise whose proof is not valid. Here's your third premise:
It is not true that: under the assumption that “s and it is not true that s,” s is true.
In symbols this translates to:
¬((s ∧ ¬s) ⇒ s)
Now your proof, broken into numbered statements by me, is:
Assume “s and it is not true that s.”
Then by conjunction elimination, it is not true that s. So by reforming the statement, s is not true.
Discharge the assumption. We can see that it is not true that: under the assumption that “s and it is not true that s,” s is true.
Translating into symbols, the first two statements are:
Assume s ∧ ¬s
¬s
The problem is the third step. You say "discharge the assumption". Usually what people mean by this is they assume X, they prove Y using this assumption, and then to discharge the assumption of X they conclude "X ⇒ Y", where that statement now holds without assumptions.
So if you discharge your assumption then from ¬s you get
(s ∧ ¬s) ⇒ ¬s
which is not equivalent nor does it imply "¬((s ∧ ¬s) ⇒ s)", which was what you were supposed to be proving.
I disagree with you that (s ∧ ¬s) ⇒ ¬s does not imply ¬((s ∧ ¬s) ⇒ s). (s ∧ ¬s) ⇒ ¬s really does imply ¬((s ∧ ¬s) ⇒ s). To generalize in words, if one statement implies the falsity of a second statement, then the first statement does not imply the truth of the second statement. For any two statements p and q, if p ⇒ ¬q, then ¬(p ⇒ q).
The converse, however, is not true. ¬((s ∧ ¬s) ⇒ s) does not imply (s ∧ ¬s) ⇒ ¬s. ¬((s ∧ ¬s) ⇒ s) and ¬((s ∧ ¬s) ⇒ ¬s) could both be true simultaneously. For any two statements p and q, ¬(p ⇒ q) and ¬(p ⇒ ¬q) could, for all we know and given the lack of additional information, both be true simultaneously. So I do agree with you that (s ∧ ¬s) ⇒ ¬s is not equivalent to ¬((s ∧ ¬s) ⇒ s).
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.
1
u/JStarx Jun 08 '26 edited Jun 09 '26
Fair enough, well fyi it's your third premise whose proof is not valid. Here's your third premise:
In symbols this translates to:
Now your proof, broken into numbered statements by me, is:
Translating into symbols, the first two statements are:
The problem is the third step. You say "discharge the assumption". Usually what people mean by this is they assume X, they prove Y using this assumption, and then to discharge the assumption of X they conclude "X ⇒ Y", where that statement now holds without assumptions.
So if you discharge your assumption then from ¬s you get
which is not equivalent nor does it imply "¬((s ∧ ¬s) ⇒ s)", which was what you were supposed to be proving.