Atario wrote:
Ah, but dear Prof, is not "canon is restricted to the game itself"
also just a dev statement, and thus equally dismissible?
Oh dear, I've gone cross-eyed.
There's a catch. "All dev statements are canon" is the proposed axiom. If I wish to disagree with this axiom, I need only prove a negation (i.e., "Not all dev statements are canon"), not the inversion (i.e., "All dev statements are not canon").
So, let's turn this into symbolic logic (for the people who are more math-oriented):
We will call the proposed Axiom "A" because we're lazy like that. Similarly, we will identify "D" as the set containing all dev statements. Finally, "C" is the set containing everything that is canon (or, for the sake of logic, all statements that are true).
"A" can be defined as "x ∈ D ⇒ x ∈ C" or "If 'x' is a dev statement, then 'x' is canon." or (for the ease of writing) "All dev statements are canon.'"
"C" can be defined as "x ⇔ x ∈ C" or "'x' is true if and only if 'x' is canon."
To hold "A" as axiomatic, "A ∈ C ⇔ A" or "'All dev statements are canon' is canon if and only if 'All dev statements are canon' is true."
And then we throw the wrench in the gears, because of the following:
S ∈ D - "'Canon is restricted to the game itself' is a dev statement."
and
S ⇒ ¬A - "If 'Canon is restricted to the game itself' then ''Not all dev statements are canon.'"
We run that through the other two propositions and we end up with this:
S ∈ D ⇒ ¬A ∈ C - "If 'Canon is restricted to the game itself' is a dev statement, then 'Not all dev statements are canon' is canon."
The definition of negation is that "x ⇔ ¬(¬x)" - "'x' is true if and only if 'not x' is false." This means that we have the following property for "C":
x ∈ C ⇔ ¬(¬x ∈ C) - "'x' is canon if and only if 'not x' is not canon."
So we then have:
¬A ∈ C ⇔ ¬[¬(¬A) ∈ C] - simplified to ¬A ∈ C ⇔ ¬(A ∈ C) - "'Not all dev statements are canon' is canon if and only if 'All dev statements are canon' is not canon."
And the contrapositive:
¬(¬A ∈ C) ⇔ ¬{¬[¬(¬A) ∈ C]} - simplified to A ∈ C ⇔ ¬(¬A ∈ C) - "'All dev statements are canon' is canon if and only if 'Not all dev statements are canon' is not canon."
But our logical extension has resulted in the following:
A ∈ C ⇔ ¬(¬A ∈ C) - "If 'All dev statements are canon' is canon, then 'Not all dev statements are canon' is canon."
From those three propositions (¬A ∈ C ⇔ ¬(A ∈ C), A ∈ C ⇔ ¬(¬A ∈ C), A ∈ C ⇔ ¬(¬A ∈ C)), there is one formulation that avoids contradiction:
¬A ∈ C - "'Not all dev statements are canon' is canon."
Which means, by extension:
¬(A ∈ C) - "'All dev statements are canon' is not canon."
and
¬(A ∈ C) ⇔ ¬A - "'All dev statements are canon' is not canon if and only if 'All dev statements are canon' is not true."
Therefore, "All dev statements are canon" is not true.
NOW you have permission to go cross-eyed.
