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Theorem sblbis 1850
 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 1849 . 2 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2 sblbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi2i 220 . 2 (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
41, 3bitri 177 1 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
 Colors of variables: wff set class Syntax hints:   ↔ wb 102  [wsb 1661 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 640  ax-5 1352  ax-7 1353  ax-gen 1354  ax-ie1 1398  ax-ie2 1399  ax-8 1411  ax-10 1412  ax-11 1413  ax-i12 1414  ax-4 1416  ax-17 1435  ax-i9 1439  ax-ial 1443  ax-i5r 1444 This theorem depends on definitions:  df-bi 114  df-nf 1366  df-sb 1662 This theorem is referenced by:  sb8eu  1929  sb8euh  1939  sb8iota  4902
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