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Theorem sblbis 1882
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 1881 . 2 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2 sblbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi2i 225 . 2 (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
41, 3bitri 182 1 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
Colors of variables: wff set class
Syntax hints:  wb 103  [wsb 1692
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473
This theorem depends on definitions:  df-bi 115  df-nf 1395  df-sb 1693
This theorem is referenced by:  sb8eu  1961  sb8euh  1971  sb8iota  4987
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