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Theorem sbeqi 35911
 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 2078 . 2 (∀𝑧(𝜑𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2 sbequ 2088 . 2 (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
31, 2sylan9bbr 514 1 ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 209   ∧ wa 399  ∀wal 1536   = wceq 1538  [wsb 2069 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 This theorem depends on definitions:  df-bi 210  df-an 400  df-ex 1782  df-sb 2070 This theorem is referenced by: (None)
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