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Theorem sblbis 1831
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 1830 . 2 ([y / x](χφ) ↔ ([y / x]χ ↔ [y / x]φ))
2 sblbis.1 . . 3 ([y / x]φψ)
32bibi2i 216 . 2 (([y / x]χ ↔ [y / x]φ) ↔ ([y / x]χψ))
41, 3bitri 173 1 ([y / x](χφ) ↔ ([y / x]χψ))
Colors of variables: wff set class
Syntax hints:  wb 98  [wsb 1642
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425
This theorem depends on definitions:  df-bi 110  df-nf 1347  df-sb 1643
This theorem is referenced by:  sb8eu  1910  sb8euh  1920  sb8iota  4817
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