Theorem spsbbi 1725

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 1724	. . 3 ⊢ (∀𝑥(𝜑 → 𝜓) → ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
2		spsbim 1724	. . 3 ⊢ (∀𝑥(𝜓 → 𝜑) → ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
3	1, 2	anim12i 321	. 2 ⊢ ((∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)) → (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
4		albiim 1376	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) ↔ (∀𝑥(𝜑 → 𝜓) ∧ ∀𝑥(𝜓 → 𝜑)))
5		dfbi2 368	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6	3, 4, 5	3imtr4i 190	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 97   ↔ wb 98  ∀wal 1241  [wsb 1645

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1336  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-4 1400  ax-ial 1427

This theorem depends on definitions:  df-bi 110  df-sb 1646

This theorem is referenced by:  sbbid  1726  relelfvdm  5205