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Theorem spv 2400
 Description: Specialization, using implicit substitution. Usage of this theorem is discouraged because it depends on ax-13 2379. Use the weaker spvv 2003 if possible. (Contributed by NM, 30-Aug-1993.) (New usage is discouraged.)
Hypothesis
Ref Expression
spv.1 (𝑥 = 𝑦 → (𝜑𝜓))
Assertion
Ref Expression
spv (∀𝑥𝜑𝜓)
Distinct variable group:   𝜓,𝑥
Allowed substitution hints:   𝜑(𝑥,𝑦)   𝜓(𝑦)

Proof of Theorem spv
StepHypRef Expression
1 spv.1 . . 3 (𝑥 = 𝑦 → (𝜑𝜓))
21biimpd 232 . 2 (𝑥 = 𝑦 → (𝜑𝜓))
32spimv 2397 1 (∀𝑥𝜑𝜓)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209  ∀wal 1536 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-12 2175  ax-13 2379 This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-nf 1786 This theorem is referenced by:  cbvalvOLD  2409  axc11n-16  36250
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