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Theorem sbeqi 36430
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 2075 . 2 (∀𝑧(𝜑𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2 sbequ 2085 . 2 (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
31, 2sylan9bbr 511 1 ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 205  wa 396  wal 1538   = wceq 1540  [wsb 2066
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1912  ax-6 1970  ax-7 2010
This theorem depends on definitions:  df-bi 206  df-an 397  df-ex 1781  df-sb 2067
This theorem is referenced by: (None)
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