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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 | |||||||||||||

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| 7. | ¬¬B | ¬Intro: 4-6 | ||||||||||||

| 8. | B | ¬Elim: 7 |

Assume that for all sets A, A ⊬ ⊥ entails A ⊭_{TT} ⊥

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 A ⊭_{TT} ⊥

then A ⊨_{TT} B entails A ⊢ B