Theorem sbbi 2296

Description: Equivalence inside and outside of a substitution are equivalent. (Contributed by NM, 14-May-1993.)

Assertion

Ref	Expression
sbbi	⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Proof of Theorem sbbi

Step	Hyp	Ref	Expression
1		dfbi2 474	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2	1	sbbii 2071	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ [𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
3		sbim 2291	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
4		sbim 2291	. . . 4 ⊢ ([𝑦 / 𝑥](𝜓 → 𝜑) ↔ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
5	3, 4	anbi12i 626	. . 3 ⊢ (([𝑦 / 𝑥](𝜑 → 𝜓) ∧ [𝑦 / 𝑥](𝜓 → 𝜑)) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6		sban 2075	. . 3 ⊢ ([𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ ([𝑦 / 𝑥](𝜑 → 𝜓) ∧ [𝑦 / 𝑥](𝜓 → 𝜑)))
7		dfbi2 474	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
8	5, 6, 7	3bitr4i 303	. 2 ⊢ ([𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))
9	2, 8	bitri 275	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 395  [wsb 2059

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-10 2129  ax-12 2163

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 845  df-ex 1774  df-nf 1778  df-sb 2060

This theorem is referenced by:  sblbis  2297  sbrbis  2298  pm13.183  3649  sbcbig  3824  sb8iota  6498  bj-sbidmOLD  36220