Theorem sbeqi 37871

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 2069	. 2 ⊢ (∀𝑧(𝜑 ↔ 𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2		sbequ 2079	. 2 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
3	1, 2	sylan9bbr 509	1 ⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))

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)