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Theorem exan 1599
Description: Place a conjunct in the scope of an existential quantifier. (Contributed by NM, 18-Aug-1993.) (Proof shortened by Andrew Salmon, 25-May-2011.)
Hypothesis
Ref Expression
exan.1 (∃𝑥𝜑𝜓)
Assertion
Ref Expression
exan 𝑥(𝜑𝜓)

Proof of Theorem exan
StepHypRef Expression
1 hbe1 1400 . . . 4 (∃𝑥𝜑 → ∀𝑥𝑥𝜑)
2119.28h 1470 . . 3 (∀𝑥(∃𝑥𝜑𝜓) ↔ (∃𝑥𝜑 ∧ ∀𝑥𝜓))
3 exan.1 . . 3 (∃𝑥𝜑𝜓)
42, 3mpgbi 1357 . 2 (∃𝑥𝜑 ∧ ∀𝑥𝜓)
5 19.29r 1528 . 2 ((∃𝑥𝜑 ∧ ∀𝑥𝜓) → ∃𝑥(𝜑𝜓))
64, 5ax-mp 7 1 𝑥(𝜑𝜓)
Colors of variables: wff set class
Syntax hints:  wa 101  wal 1257  wex 1397
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-4 1416  ax-ial 1443
This theorem depends on definitions:  df-bi 114
This theorem is referenced by:  bm1.3ii  3906  bdbm1.3ii  10398
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