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Theorem sbrbif 2074
 Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.) (Revised by Mario Carneiro, 4-Oct-2016.)
Hypotheses
Ref Expression
sbrbif.1 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 2073 . 2 ([y / x](φχ) ↔ (ψ ↔ [y / x]χ))
3 sbrbif.1 . . . 4 xχ
43sbf 2026 . . 3 ([y / x]χχ)
54bibi2i 304 . 2 ((ψ ↔ [y / x]χ) ↔ (ψχ))
62, 5bitri 240 1 ([y / x](φχ) ↔ (ψχ))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176  Ⅎwnf 1544  [wsb 1648 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 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: (None)
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