Theorem hbor 1478

Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑 ∨ 𝜓). (Contributed by NM, 5-Aug-1993.) (Revised by NM, 2-Feb-2015.)

Hypotheses

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

Assertion

Ref	Expression
hbor	⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))

Proof of Theorem hbor

Step	Hyp	Ref	Expression
1		hb.1	. . 3 ⊢ (𝜑 → ∀𝑥𝜑)
2		orc 665	. . . 4 ⊢ (𝜑 → (𝜑 ∨ 𝜓))
3	2	alimi 1384	. . 3 ⊢ (∀𝑥𝜑 → ∀𝑥(𝜑 ∨ 𝜓))
4	1, 3	syl 14	. 2 ⊢ (𝜑 → ∀𝑥(𝜑 ∨ 𝜓))
5		hb.2	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
6		olc 664	. . . 4 ⊢ (𝜓 → (𝜑 ∨ 𝜓))
7	6	alimi 1384	. . 3 ⊢ (∀𝑥𝜓 → ∀𝑥(𝜑 ∨ 𝜓))
8	5, 7	syl 14	. 2 ⊢ (𝜓 → ∀𝑥(𝜑 ∨ 𝜓))
9	4, 8	jaoi 668	1 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))

Syntax hints:   → wi 4   ∨ wo 661  ∀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

This theorem is referenced by:  hb3or  1481  nfor  1506  19.43  1559