Theorem 3bior1fd 1505

Description: A disjunction is equivalent to a threefold disjunction with single falsehood, analogous to biorf 949. (Contributed by Alexander van der Vekens, 8-Sep-2017.)

Hypothesis

Ref	Expression
3biorfd.1	⊢ (𝜑 → ¬ 𝜃)

Assertion

Ref	Expression
3bior1fd	⊢ (𝜑 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ 𝜒 ∨ 𝜓)))

Proof of Theorem 3bior1fd

Step	Hyp	Ref	Expression
1		3biorfd.1	. . 3 ⊢ (𝜑 → ¬ 𝜃)
2		biorf 949	. . 3 ⊢ (¬ 𝜃 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ (𝜒 ∨ 𝜓))))
3	1, 2	syl 18	. 2 ⊢ (𝜑 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ (𝜒 ∨ 𝜓))))
4		3orass 1105	. 2 ⊢ ((𝜃 ∨ 𝜒 ∨ 𝜓) ↔ (𝜃 ∨ (𝜒 ∨ 𝜓)))
5	3, 4	bitr4di 292	1 ⊢ (𝜑 → ((𝜒 ∨ 𝜓) ↔ (𝜃 ∨ 𝜒 ∨ 𝜓)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∨ wo 860   ∨ w3o 1101

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 861  df-3or 1103

This theorem is used by:  3bior1fand  1506  3bior2fd  1507  nb3grprlem2  29742