Theorem ts3or1 39065

Description: A Tseitin axiom for triple logical disjunction, in deduction form. (Contributed by Giovanni Mascellani, 25-Mar-2018.)

Assertion

Ref	Expression
ts3or1	⊢ (𝜃 → (((𝜑 ∨ 𝜓) ∨ 𝜒) ∨ ¬ (𝜑 ∨ 𝜓 ∨ 𝜒)))

Proof of Theorem ts3or1

Step	Hyp	Ref	Expression
1		exmidd 909	. 2 ⊢ (𝜃 → (((𝜑 ∨ 𝜓) ∨ 𝜒) ∨ ¬ ((𝜑 ∨ 𝜓) ∨ 𝜒)))
2		df-3or 1104	. . . 4 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
3	2	notbii 323	. . 3 ⊢ (¬ (𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ¬ ((𝜑 ∨ 𝜓) ∨ 𝜒))
4	3	orbi2i 926	. 2 ⊢ ((((𝜑 ∨ 𝜓) ∨ 𝜒) ∨ ¬ (𝜑 ∨ 𝜓 ∨ 𝜒)) ↔ (((𝜑 ∨ 𝜓) ∨ 𝜒) ∨ ¬ ((𝜑 ∨ 𝜓) ∨ 𝜒)))
5	1, 4	sylibr 237	1 ⊢ (𝜃 → (((𝜑 ∨ 𝜓) ∨ 𝜒) ∨ ¬ (𝜑 ∨ 𝜓 ∨ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∨ wo 861   ∨ w3o 1102

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-or 862  df-3or 1104

This theorem is used by: (None)