Theorem sbeqi 38119

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 2073	. 2 ⊢ (∀𝑧(𝜑 ↔ 𝜓) → ([𝑥 / 𝑧]𝜑 ↔ [𝑥 / 𝑧]𝜓))
2		sbequ 2083	. 2 ⊢ (𝑥 = 𝑦 → ([𝑥 / 𝑧]𝜓 ↔ [𝑦 / 𝑧]𝜓))
3	1, 2	sylan9bbr 510	1 ⊢ ((𝑥 = 𝑦 ∧ ∀𝑧(𝜑 ↔ 𝜓)) → ([𝑥 / 𝑧]𝜑 ↔ [𝑦 / 𝑧]𝜓))

Syntax hints:   → wi 4   ↔ wb 206   ∧ wa 395  ∀wal 1535   = wceq 1537  [wsb 2064

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1793  ax-4 1807  ax-5 1909  ax-6 1967  ax-7 2007

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1778  df-sb 2065

This theorem is referenced by: (None)