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Theorem sblbis 2072
 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 2071 . 2 ([y / x](χφ) ↔ ([y / x]χ ↔ [y / x]φ))
2 sblbis.1 . . 3 ([y / x]φψ)
32bibi2i 304 . 2 (([y / x]χ ↔ [y / x]φ) ↔ ([y / x]χψ))
41, 3bitri 240 1 ([y / x](χφ) ↔ ([y / x]χψ))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176  [wsb 1648 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925 This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649 This theorem is referenced by:  sb8eu  2222  sb8iota  4346
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