Theorem sbbi 1933

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 386	. . 3 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
2	1	sbbii 1739	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ [𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)))
3		sbim 1927	. . . 4 ⊢ ([𝑦 / 𝑥](𝜑 → 𝜓) ↔ ([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓))
4		sbim 1927	. . . 4 ⊢ ([𝑦 / 𝑥](𝜓 → 𝜑) ↔ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑))
5	3, 4	anbi12i 456	. . 3 ⊢ (([𝑦 / 𝑥](𝜑 → 𝜓) ∧ [𝑦 / 𝑥](𝜓 → 𝜑)) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
6		sban 1929	. . 3 ⊢ ([𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ ([𝑦 / 𝑥](𝜑 → 𝜓) ∧ [𝑦 / 𝑥](𝜓 → 𝜑)))
7		dfbi2 386	. . 3 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓) ↔ (([𝑦 / 𝑥]𝜑 → [𝑦 / 𝑥]𝜓) ∧ ([𝑦 / 𝑥]𝜓 → [𝑦 / 𝑥]𝜑)))
8	5, 6, 7	3bitr4i 211	. 2 ⊢ ([𝑦 / 𝑥]((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))
9	2, 8	bitri 183	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜓))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104  [wsb 1736

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-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516

This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737

This theorem is referenced by:  sblbis  1934  sbrbis  1935  sbco  1942  sbcocom  1944  elsb3  1952  elsb4  1953  sb8eu  2013  sb8euh  2023  pm13.183  2826  sbcbig  2959  sb8iota  5103