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Theorem sbeqi 34881
 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 2024 . 2 (∀𝑧(𝜑𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2 sbequ 2035 . 2 (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
31, 2sylan9bbr 503 1 ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 387  ∀wal 1505   = wceq 1507  [wsb 2015 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1758  ax-4 1772  ax-5 1869  ax-6 1928  ax-7 1965 This theorem depends on definitions:  df-bi 199  df-an 388  df-ex 1743  df-sb 2016 This theorem is referenced by: (None)
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