Is this statement false? If I've done the truth table right, then it means that the first line of the proof is wrong. — Brendan Golledge
If definitions aren't subject to truth apt, then can I say, "Let 'X' mean a married bachelor," and that this sentence is not truth apt? — Brendan Golledge
Michael said earlier that a definition is not truth apt. I can see how that would be the case if you defined an entirely new variable, such as Z <-> (X -> Y). However, since you are setting X equal to itself, you can do a truth table on it. — Brendan Golledge
I don't think you can get away with any arbitrary definition. — Brendan Golledge
The way I see it, these paradoxes show in a nice way how all truth is an idealization. — Apustimelogist
To add to the above:
If X := X->Y then X <-> (X->Y).
But we don' t have the converse that if X <-> (X->Y) then X := X->Y.
So X := X->Y is not equivalent with X <-> (X->Y).
So we can't dispense the paradox by incorrectly saying that it reduces to X <-> (X -> Y). — TonesInDeepFreeze
Then you get X := NOT X — Brendan Golledge
You can use a truth table to prove NOT X <-> (X -> F).
(X -> Y) <-> (X -> F) in the case where Y is false, so this applies to Curry's paradox as well as "this sentence is false". — Brendan Golledge
Then you take your definition X := (X ->F) and substitute NOT X for the second part. — Brendan Golledge
Trouble is, that's just an idealisation. — Banno
There are true sentences — Banno
Are you saying it is better to play the gamein the wrong way?...when you agree to play the game in the right way. — Apustimelogist
You can use a truth table to prove NOT X <-> (X -> F).
(X -> Y) <-> (X -> F) in the case where Y is false, so this applies to Curry's paradox as well as "this sentence is false".
— Brendan Golledge
Yes:
|- ~X <-> (X -> F)
If Y is false then (X -> Y) <-> (X -> F) is true.
That's not Curry's paradox.
Then you take your definition X := (X ->F) and substitute NOT X for the second part.
— Brendan Golledge
Who does that? You? Did someone previously define?:
X := (X -> F) — TonesInDeepFreeze
To add to the above:
If X := X->Y then X <-> (X->Y).
But we don' t have the converse that if X <-> (X->Y) then X := X->Y.
So X := X->Y is not equivalent with X <-> (X->Y).
So we can't dispense the paradox by incorrectly saying that it reduces to X <-> (X -> Y). — TonesInDeepFreeze
Who does that? You? Did someone previously define?:
X := (X -> F) — TonesInDeepFreeze
That's not Curry's paradox. — TonesInDeepFreeze
If X := X->Y then X <-> (X->Y). — TonesInDeepFreeze
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