Theorem sbrbif 2015

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)

Hypotheses

Ref	Expression
sbrbif.1	⊢ (𝜒 → ∀𝑥𝜒)
sbrbif.2	⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)

Assertion

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

Proof of Theorem sbrbif

Step	Hyp	Ref	Expression
1		sbrbif.2	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
2	1	sbrbis 2014	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
3		sbrbif.1	. . . 4 ⊢ (𝜒 → ∀𝑥𝜒)
4	3	sbh 1824	. . 3 ⊢ ([𝑦 / 𝑥]𝜒 ↔ 𝜒)
5	4	bibi2i 227	. 2 ⊢ ((𝜓 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ 𝜒))
6	2, 5	bitri 184	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1395  [wsb 1810

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583

This theorem depends on definitions:  df-bi 117  df-nf 1509  df-sb 1811

This theorem is referenced by: (None)