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Theorem hbor 1526
Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑𝜓). (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 702 . . . 4 (𝜑 → (𝜑𝜓))
32alimi 1435 . . 3 (∀𝑥𝜑 → ∀𝑥(𝜑𝜓))
41, 3syl 14 . 2 (𝜑 → ∀𝑥(𝜑𝜓))
5 hb.2 . . 3 (𝜓 → ∀𝑥𝜓)
6 olc 701 . . . 4 (𝜓 → (𝜑𝜓))
76alimi 1435 . . 3 (∀𝑥𝜓 → ∀𝑥(𝜑𝜓))
85, 7syl 14 . 2 (𝜓 → ∀𝑥(𝜑𝜓))
94, 8jaoi 706 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wo 698  wal 1333
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-io 699  ax-5 1427  ax-gen 1429
This theorem depends on definitions:  df-bi 116
This theorem is referenced by:  hb3or  1529  nfor  1554  19.43  1608
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