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Theorem sblbis 1911
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 1910 . 2 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑))
2 sblbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi2i 226 . 2 (([𝑦 / 𝑥]𝜒 ↔ [𝑦 / 𝑥]𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
41, 3bitri 183 1 ([𝑦 / 𝑥](𝜒𝜑) ↔ ([𝑦 / 𝑥]𝜒𝜓))
Colors of variables: wff set class
Syntax hints:  wb 104  [wsb 1720
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 683  ax-5 1408  ax-7 1409  ax-gen 1410  ax-ie1 1454  ax-ie2 1455  ax-8 1467  ax-10 1468  ax-11 1469  ax-i12 1470  ax-4 1472  ax-17 1491  ax-i9 1495  ax-ial 1499  ax-i5r 1500
This theorem depends on definitions:  df-bi 116  df-nf 1422  df-sb 1721
This theorem is referenced by:  sb8eu  1990  sb8euh  2000  sb8iota  5065
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