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Theorem sbeqi 32938
Description: Equality deduction for substitution. (Contributed by Giovanni Mascellani, 10-Apr-2018.)
Assertion
Ref Expression
sbeqi ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))

Proof of Theorem sbeqi
StepHypRef Expression
1 spsbbi 2386 . 2 (∀𝑧(𝜑𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2 sbequ 2360 . 2 (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
31, 2sylan9bbr 732 1 ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 194  wa 382  wal 1472   = wceq 1474  [wsb 1866
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-12 2032  ax-13 2229
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867
This theorem is referenced by: (None)
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