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Theorem sbeqi 37871
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 2069 . 2 (∀𝑧(𝜑𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2 sbequ 2079 . 2 (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
31, 2sylan9bbr 509 1 ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 394  wal 1532   = wceq 1534  [wsb 2060
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004
This theorem depends on definitions:  df-bi 206  df-an 395  df-ex 1775  df-sb 2061
This theorem is referenced by: (None)
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