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Theorem sbrbif 1833
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypotheses
Ref Expression
sbrbif.1 (χxχ)
sbrbif.2 ([y / x]φψ)
Assertion
Ref Expression
sbrbif ([y / x](φχ) ↔ (ψχ))

Proof of Theorem sbrbif
StepHypRef Expression
1 sbrbif.2 . . 3 ([y / x]φψ)
21sbrbis 1832 . 2 ([y / x](φχ) ↔ (ψ ↔ [y / x]χ))
3 sbrbif.1 . . . 4 (χxχ)
43sbh 1656 . . 3 ([y / x]χχ)
54bibi2i 216 . 2 ((ψ ↔ [y / x]χ) ↔ (ψχ))
62, 5bitri 173 1 ([y / x](φχ) ↔ (ψχ))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 98  wal 1240  [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: (None)
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