Theorem spsbbi 1765

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 1764	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
2		spsbim 1764	. . 3 ⊢ (∀𝑥(𝜓 → 𝜑) → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
3	1, 2	anim12i 331	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)) → (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
4		albiim 1416	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)))
5		dfbi2 380	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6	3, 4, 5	3imtr4i 199	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103  ∀wal 1282  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1376  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-4 1440  ax-ial 1467

This theorem depends on definitions:  df-bi 115  df-sb 1686

This theorem is referenced by:  sbbidh  1766  sbbid  1767  relelfvdm  5226