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Theorem hbim 1524
 Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑 → 𝜓). (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 1487 . . 3 (∀𝑥𝜑𝜑)
2 hb.2 . . 3 (𝜓 → ∀𝑥𝜓)
31, 2imim12i 59 . 2 ((𝜑𝜓) → (∀𝑥𝜑 → ∀𝑥𝜓))
4 ax-i5r 1515 . 2 ((∀𝑥𝜑 → ∀𝑥𝜓) → ∀𝑥(∀𝑥𝜑𝜓))
5 hb.1 . . . 4 (𝜑 → ∀𝑥𝜑)
65imim1i 60 . . 3 ((∀𝑥𝜑𝜓) → (𝜑𝜓))
76alimi 1431 . 2 (∀𝑥(∀𝑥𝜑𝜓) → ∀𝑥(𝜑𝜓))
83, 4, 73syl 17 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
 Colors of variables: wff set class Syntax hints:   → wi 4  ∀wal 1329 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-5 1423  ax-gen 1425  ax-4 1487  ax-i5r 1515 This theorem is referenced by:  hbbi  1527  hbia1  1531  19.21h  1536  19.38  1654  hbsbv  1914  hbmo1  2037  hbmo  2038  moexexdc  2083  2eu4  2092  cleqh  2239  hbral  2464
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