Proof 20200424

In a previous post, I formulated an argument based on a theorem that if ¬◊¬(φ → ψ) and ◊(φ ∧ χ) are provable, then ◊(ψ ∧ χ) is provable. In this post, I prove the theorem.

THEOREM If ¬◊¬(φ → ψ) and ◊(φ ∧ χ) are provable, then ◊(ψ ∧ χ) is provable.

1. ¬◊¬(φ → ψ)
2. ◊(φ ∧ χ)
3. ◊φ *
4. ◊χ *
5. ◊φ → ψ **
6. ψ
7. |¬◊(ψ ∧ χ)
8. |¬◊¬(χ → ¬ψ) **
9. |◊χ → ¬ψ
10. |¬ψ
11. |ψ ∧ ¬ψ
12. ◊(ψ ∧ χ)

LEMMA 1 If ◊(φ ∧ χ) is provable, then ◊φ is provable.

1. ◊(φ ∧ χ)
2. |¬◊φ
3. |¬◊¬¬φ
4. |¬◊¬(¬φ ∨ ¬χ) ***
5. |¬◊(φ ∧ χ)
6. |◊(φ ∧ χ) ∧ ¬◊(φ ∧ χ)
7. ◊φ

LEMMA 2 If ¬◊¬(φ → ψ) is provable, then ◊φ → ψ is provable.

1. ¬◊¬(φ → ψ)
2. ¬◊¬(¬φ v ψ) ****
3. ¬◊(φ ∧ ¬ψ) *****
4. ¬(◊φ ∧ ◊¬ψ) *
5. ¬◊φ ∨ ¬◊¬ψ
6. |◊φ
7. |¬¬◊φ
8. |¬◊¬ψ
9. |ψ
10. ◊φ → ψ

Notes. * Assuming LEMMA 1. ** Assuming LEMMA 2. *** According to S5 AS2, because ¬◊¬ (¬φ → (¬φ ∨ ¬χ)), therefore ¬◊¬¬φ → ¬◊¬(¬φ ∨ ¬χ). **** Material Implication. ***** De Morgan’s Theorems.

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