Theorem hbor 1438

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 633	. . . 4 ⊢ (𝜑 → (𝜑 ∨ 𝜓))
3	2	alimi 1344	. . 3 ⊢ (∀𝑥𝜑 → ∀𝑥(𝜑 ∨ 𝜓))
4	1, 3	syl 14	. 2 ⊢ (𝜑 → ∀𝑥(𝜑 ∨ 𝜓))
5		hb.2	. . 3 ⊢ (𝜓 → ∀𝑥𝜓)
6		olc 632	. . . 4 ⊢ (𝜓 → (𝜑 ∨ 𝜓))
7	6	alimi 1344	. . 3 ⊢ (∀𝑥𝜓 → ∀𝑥(𝜑 ∨ 𝜓))
8	5, 7	syl 14	. 2 ⊢ (𝜓 → ∀𝑥(𝜑 ∨ 𝜓))
9	4, 8	jaoi 636	1 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑥(𝜑 ∨ 𝜓))

Syntax hints:   → wi 4   ∨ wo 629  ∀wal 1241

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-gen 1338

This theorem depends on definitions:  df-bi 110

This theorem is referenced by:  hb3or  1441  nfor  1466  19.43  1519