Theorem spsbbi 1816

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 1815	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
2		spsbim 1815	. . 3 ⊢ (∀𝑥(𝜓 → 𝜑) → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
3	1, 2	anim12i 336	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)) → (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
4		albiim 1463	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)))
5		dfbi2 385	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6	3, 4, 5	3imtr4i 200	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  ∀wal 1329  [wsb 1735

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1423  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-4 1487  ax-ial 1514

This theorem depends on definitions:  df-bi 116  df-sb 1736

This theorem is referenced by:  sbbidh  1817  sbbid  1818  relelfvdm  5453