Theorem sbeqi 33968

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 2402	. 2 ⊢ (∀𝑧(𝜑 ↔ 𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2		sbequ 2376	. 2 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
3	1, 2	sylan9bbr 737	1 ⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))

Syntax hints:   → wi 4   ↔ wb 196   ∧ wa 384  ∀wal 1481   = wceq 1483  [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by: (None)