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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 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))

Proof of Theorem hbor
StepHypRef Expression
1 hb.1 . . 3 (𝜑 → ∀𝑥𝜑)
2 orc 633 . . . 4 (𝜑 → (𝜑𝜓))
32alimi 1344 . . 3 (∀𝑥𝜑 → ∀𝑥(𝜑𝜓))
41, 3syl 14 . 2 (𝜑 → ∀𝑥(𝜑𝜓))
5 hb.2 . . 3 (𝜓 → ∀𝑥𝜓)
6 olc 632 . . . 4 (𝜓 → (𝜑𝜓))
76alimi 1344 . . 3 (∀𝑥𝜓 → ∀𝑥(𝜑𝜓))
85, 7syl 14 . 2 (𝜓 → ∀𝑥(𝜑𝜓))
94, 8jaoi 636 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
 Colors of variables: wff set class 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
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