Theorem hb3or 1481

Description: If 𝑥 is not free in 𝜑, 𝜓, and 𝜒, it is not free in (𝜑 ∨ 𝜓 ∨ 𝜒). (Contributed by NM, 14-Sep-2003.)

Hypotheses

Ref	Expression
hb.1	⊢ (𝜑 → ∀𝑥𝜑)
hb.2	⊢ (𝜓 → ∀𝑥𝜓)
hb.3	⊢ (𝜒 → ∀𝑥𝜒)

Assertion

Ref	Expression
hb3or	⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → ∀𝑥(𝜑 ∨ 𝜓 ∨ 𝜒))

Proof of Theorem hb3or

Step	Hyp	Ref	Expression
1		df-3or 920	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒))
2		hb.1	. . . 4 ⊢ (𝜑 → ∀𝑥𝜑)
3		hb.2	. . . 4 ⊢ (𝜓 → ∀𝑥𝜓)
4	2, 3	hbor 1478	. . 3 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))
5		hb.3	. . 3 ⊢ (𝜒 → ∀𝑥𝜒)
6	4, 5	hbor 1478	. 2 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) → ∀𝑥((𝜑 ∨ 𝜓) ∨ 𝜒))
7	1, 6	hbxfrbi 1401	1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → ∀𝑥(𝜑 ∨ 𝜓 ∨ 𝜒))

Syntax hints:   → wi 4   ∨ wo 661   ∨ w3o 918  ∀wal 1282

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-gen 1378

This theorem depends on definitions:  df-bi 115  df-3or 920

This theorem is referenced by: (None)