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Theorem hbim 1482
Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑𝜓). (Contributed by NM, 5-Aug-1993.) (Proof shortened by O'Cat, 3-Mar-2008.) (Revised by Mario Carneiro, 2-Feb-2015.)
Hypotheses
Ref Expression
hb.1 (𝜑 → ∀𝑥𝜑)
hb.2 (𝜓 → ∀𝑥𝜓)
Assertion
Ref Expression
hbim ((𝜑𝜓) → ∀𝑥(𝜑𝜓))

Proof of Theorem hbim
StepHypRef Expression
1 ax-4 1445 . . 3 (∀𝑥𝜑𝜑)
2 hb.2 . . 3 (𝜓 → ∀𝑥𝜓)
31, 2imim12i 58 . 2 ((𝜑𝜓) → (∀𝑥𝜑 → ∀𝑥𝜓))
4 ax-i5r 1473 . 2 ((∀𝑥𝜑 → ∀𝑥𝜓) → ∀𝑥(∀𝑥𝜑𝜓))
5 hb.1 . . . 4 (𝜑 → ∀𝑥𝜑)
65imim1i 59 . . 3 ((∀𝑥𝜑𝜓) → (𝜑𝜓))
76alimi 1389 . 2 (∀𝑥(∀𝑥𝜑𝜓) → ∀𝑥(𝜑𝜓))
83, 4, 73syl 17 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1287
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-5 1381  ax-gen 1383  ax-4 1445  ax-i5r 1473
This theorem is referenced by:  hbbi  1485  hbia1  1489  19.21h  1494  19.38  1611  hbsbv  1865  hbmo1  1986  hbmo  1987  moexexdc  2032  2eu4  2041  cleqh  2187  hbral  2407
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