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Theorem sbrbif 1949
Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypotheses
Ref Expression
sbrbif.1 (𝜒 → ∀𝑥𝜒)
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 1948 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
3 sbrbif.1 . . . 4 (𝜒 → ∀𝑥𝜒)
43sbh 1763 . . 3 ([𝑦 / 𝑥]𝜒𝜒)
54bibi2i 226 . 2 ((𝜓 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓𝜒))
62, 5bitri 183 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓𝜒))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 104  wal 1340  [wsb 1749
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 699  ax-5 1434  ax-7 1435  ax-gen 1436  ax-ie1 1480  ax-ie2 1481  ax-8 1491  ax-10 1492  ax-11 1493  ax-i12 1494  ax-4 1497  ax-17 1513  ax-i9 1517  ax-ial 1521  ax-i5r 1522
This theorem depends on definitions:  df-bi 116  df-nf 1448  df-sb 1750
This theorem is referenced by: (None)
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