Theorem spsbbi 2402

Description: Specialization of biconditional. (Contributed by NM, 2-Jun-1993.)

Assertion

Ref	Expression
spsbbi	⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Proof of Theorem spsbbi

Step	Hyp	Ref	Expression
1		stdpc4 2353	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → [𝑦 / 𝑥](𝜑 ↔ 𝜓))
2		sbbi 2401	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))
3	1, 2	sylib 208	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 196  ∀wal 1481  [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:  sbbid  2403  sbeqi  33968