Theorem sbrbif 2301

Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)

Hypotheses

Ref	Expression
sbrbif.1	⊢ Ⅎ𝑥𝜒
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 2300	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
3		sbrbif.1	. . . 4 ⊢ Ⅎ𝑥𝜒
4	3	sbf 2258	. . 3 ⊢ ([𝑦 / 𝑥]𝜒 ↔ 𝜒)
5	4	bibi2i 337	. 2 ⊢ ((𝜓 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ 𝜒))
6	2, 5	bitri 275	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ 𝜒))

Syntax hints:   ↔ wb 205  Ⅎwnf 1778  [wsb 2060

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1790  ax-4 1804  ax-5 1906  ax-6 1964  ax-7 2004  ax-10 2130  ax-12 2167

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 847  df-ex 1775  df-nf 1779  df-sb 2061

This theorem is referenced by: (None)