Theorem sbbi 1978

Description: Equivalence inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-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 388	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2	1	sbbii 1779	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ [𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
3		sbim 1972	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
4		sbim 1972	. . . 4 ⊢ ([𝑦 / 𝑥](𝜓 → 𝜑) ↔ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
5	3, 4	anbi12i 460	. . 3 ⊢ (([𝑦 / 𝑥](𝜑 → 𝜓) ∧ [𝑦 / 𝑥](𝜓 → 𝜑)) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6		sban 1974	. . 3 ⊢ ([𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ ([𝑦 / 𝑥](𝜑 → 𝜓) ∧ [𝑦 / 𝑥](𝜓 → 𝜑)))
7		dfbi2 388	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
8	5, 6, 7	3bitr4i 212	. 2 ⊢ ([𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))
9	2, 8	bitri 184	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105  [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-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549

This theorem depends on definitions:  df-bi 117  df-nf 1475  df-sb 1777

This theorem is referenced by:  sblbis  1979  sbrbis  1980  sbco  1987  sbcocom  1989  sb8eu  2058  sb8euh  2068  elsb1  2174  elsb2  2175  pm13.183  2902  sbcbig  3036  sb8iota  5226