Theorem sbeqi 35439

Description: Equality deduction for substitution. (Contributed by Giovanni Mascellani, 10-Apr-2018.)

Assertion

Ref	Expression
sbeqi	⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))

Proof of Theorem sbeqi

Step	Hyp	Ref	Expression
1		spsbbi 2078	. 2 ⊢ (∀𝑧(𝜑 ↔ 𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2		sbequ 2090	. 2 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
3	1, 2	sylan9bbr 513	1 ⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))

Syntax hints:   → wi 4   ↔ wb 208   ∧ wa 398  ∀wal 1535   = wceq 1537  [wsb 2069

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 1911  ax-6 1970  ax-7 2015

This theorem depends on definitions:  df-bi 209  df-an 399  df-ex 1781  df-sb 2070

This theorem is referenced by: (None)