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Theorem hbim 1509
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 1472 . . 3 (∀𝑥𝜑𝜑)
2 hb.2 . . 3 (𝜓 → ∀𝑥𝜓)
31, 2imim12i 59 . 2 ((𝜑𝜓) → (∀𝑥𝜑 → ∀𝑥𝜓))
4 ax-i5r 1500 . 2 ((∀𝑥𝜑 → ∀𝑥𝜓) → ∀𝑥(∀𝑥𝜑𝜓))
5 hb.1 . . . 4 (𝜑 → ∀𝑥𝜑)
65imim1i 60 . . 3 ((∀𝑥𝜑𝜓) → (𝜑𝜓))
76alimi 1416 . 2 (∀𝑥(∀𝑥𝜑𝜓) → ∀𝑥(𝜑𝜓))
83, 4, 73syl 17 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wal 1314
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-5 1408  ax-gen 1410  ax-4 1472  ax-i5r 1500
This theorem is referenced by:  hbbi  1512  hbia1  1516  19.21h  1521  19.38  1639  hbsbv  1894  hbmo1  2015  hbmo  2016  moexexdc  2061  2eu4  2070  cleqh  2217  hbral  2441
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