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## If (A ⊬ ⊥ entails A ⊭TT ⊥) then (A ⊨TT B entails A ⊢ B)

First we show that if \textbf{A}, ¬B ⊢ ⊥ then \textbf{A} ⊢ B .
(To do this, we show how to turn transform a proof of the former kind into a proof of the latter kind.)
Now assume that for all sets \textbf{A}, \textbf{A} ⊬ ⊥ entails \textbf{A} ⊭\$_{TT}\$ ⊥
Suppose that \textbf{A} ⊨TT B
Then \textbf{A}, ¬B ⊨TT ⊥
So, by our assumption, \textbf{A}, ¬B ⊢ ⊥
Therefore (by the thing we showed first) \textbf{A} ⊢ B

IfA, ¬B ⊢ ⊥ then A ⊢ B

 ​ 1 A1 ​ 2 ... ​ 3 An ​ 4 ¬B ​ 5 ... ​ 6 ⊥
1.

A1

2.

...

3.

An

 ​ 4 ¬B ​ 5 ... ​ 6 ⊥
7.¬¬B¬Intro: 4-6
8.B¬Elim: 7

Assume that for all sets A, A ⊬ ⊥ entails ATT

Suppose that A TT B

(We want to show that A ⊢ B)

Therefore: A, ¬B ⊨ TT

An argument is logically valid just if there’s no possible situation in which the premises are true and the conclusion false.

Suppose for a contradiction that A, ¬B ⊬ ⊥

Then, from the assumption, we would have A, ¬B ⊭ TT

So: A, ¬B ⊢ ⊥

Therefore (by the previous result) A ⊢ B

So we assumed this.
And derived this.

So:

If A ⊬ ⊥ entails ATT

then A TT B entails A ⊢ B