Theorem bd3or 13016

Description: A disjunction of three bounded formulas is bounded. (Contributed by BJ, 3-Oct-2019.)

Hypotheses

Ref	Expression
bd3or.1	⊢ BOUNDED 𝜑
bd3or.2	⊢ BOUNDED 𝜓
bd3or.3	⊢ BOUNDED 𝜒

Assertion

Ref	Expression
bd3or	⊢ BOUNDED (𝜑 ∨ 𝜓 ∨ 𝜒)

Proof of Theorem bd3or

Step	Hyp	Ref	Expression
1		bd3or.1	. . . 4 ⊢ BOUNDED 𝜑
2		bd3or.2	. . . 4 ⊢ BOUNDED 𝜓
3	1, 2	ax-bdor 13003	. . 3 ⊢ BOUNDED (𝜑 ∨ 𝜓)
4		bd3or.3	. . 3 ⊢ BOUNDED 𝜒
5	3, 4	ax-bdor 13003	. 2 ⊢ BOUNDED ((𝜑 ∨ 𝜓) ∨ 𝜒)
6		df-3or 963	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
7	5, 6	bd0r 13012	1 ⊢ BOUNDED (𝜑 ∨ 𝜓 ∨ 𝜒)

Syntax hints:   ∨ wo 697   ∨ w3o 961  BOUNDED wbd 12999

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-bd0 13000  ax-bdor 13003

This theorem depends on definitions:  df-bi 116  df-3or 963

This theorem is referenced by: (None)