Theorem sbrbis 2343

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

Hypothesis

Ref	Expression
sbrbis.1	⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)

Assertion

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

Proof of Theorem sbrbis

Step	Hyp	Ref	Expression
1		sbbi 2341	. 2 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2		sbrbis.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi1i 340	. 2 ⊢ (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
4	1, 3	bitri 277	1 ⊢ ([𝑦 / 𝑥](𝜑 ↔ 𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))

Syntax hints:   ↔ wb 208  [wsb 2090

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-10 2175  ax-12 2212

This theorem depends on definitions:  df-bi 209  df-an 400  df-ex 1800  df-nf 1804  df-sb 2091

This theorem is referenced by:  sbrbif  2344  sbabel  2956