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Theorem sblbis 2310
Description: Introduce left biconditional inside of a substitution. (Contributed by NM, 19-Aug-1993.)
Hypothesis
Ref Expression
sblbis.1 ([𝑦 / 𝑥]𝜑𝜓)
Assertion
Ref Expression
sblbis ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))

Proof of Theorem sblbis
StepHypRef Expression
1 sbbi 2309 . 2 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2 sblbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi2i 341 . 2 (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
41, 3bitri 278 1 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
Colors of variables: wff setvar class
Syntax hints:  wb 209  [wsb 2070
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1803  ax-4 1817  ax-5 1918  ax-6 1976  ax-7 2016  ax-10 2141  ax-12 2175
This theorem depends on definitions:  df-bi 210  df-an 400  df-or 848  df-ex 1788  df-nf 1792  df-sb 2071
This theorem is referenced by:  sbie  2505  sb8eulem  2597  sb8iota  6350  wl-sb8eut  35469
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