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Theorem sblbis 2299
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 2298 . 2 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2 sblbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi2i 337 . 2 (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
41, 3bitri 275 1 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
Colors of variables: wff setvar class
Syntax hints:  wb 205  [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:  sbie  2497  sb8eulem  2588  sbhypf  3536  sb8iota  6512  wl-sb8eut  37045  wl-sb8eutv  37046
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