Theorem sblbis 2306

Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993.)

Hypothesis

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

Assertion

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

Proof of Theorem sblbis

Step	Hyp	Ref	Expression
1		sbbi 2305	. 2 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2		sblbis.1	. . 3 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
3	2	bibi2i 338	. 2 ⊢ (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))
4	1, 3	bitri 274	1 ⊢ ([𝑦 / 𝑥](𝜒 ↔ 𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ 𝜓))

Syntax hints:   ↔ wb 205  [wsb 2067

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-10 2137  ax-12 2171

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-ex 1783  df-nf 1787  df-sb 2068

This theorem is referenced by:  sbie  2506  sb8eulem  2598  sb8iota  6403  wl-sb8eut  35732