Theorem spsbbi 1858

Description: Specialization of biconditional. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 21-Jan-2018.)

Assertion

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

Proof of Theorem spsbbi

Step	Hyp	Ref	Expression
1		spsbim 1857	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
2		spsbim 1857	. . 3 ⊢ (∀𝑥(𝜓 → 𝜑) → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
3	1, 2	anim12i 338	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)) → (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
4		albiim 1501	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)))
5		dfbi2 388	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6	3, 4, 5	3imtr4i 201	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1362  [wsb 1776

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1461  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-4 1524  ax-ial 1548

This theorem depends on definitions:  df-bi 117  df-sb 1777

This theorem is referenced by:  sbbidh  1859  sbbid  1860  sbbidv  1899  relelfvdm  5590