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Theorem sbrbis 2073
 Description: Introduce right biconditional inside of a substitution. (Contributed by NM, 18-Aug-1993.)
Hypothesis
Ref Expression
sbrbis.1 ([y / x]φψ)
Assertion
Ref Expression
sbrbis ([y / x](φχ) ↔ (ψ ↔ [y / x]χ))

Proof of Theorem sbrbis
StepHypRef Expression
1 sbbi 2071 . 2 ([y / x](φχ) ↔ ([y / x]φ ↔ [y / x]χ))
2 sbrbis.1 . . 3 ([y / x]φψ)
32bibi1i 305 . 2 (([y / x]φ ↔ [y / x]χ) ↔ (ψ ↔ [y / x]χ))
41, 3bitri 240 1 ([y / x](φχ) ↔ (ψ ↔ [y / x]χ))
 Colors of variables: wff setvar class Syntax hints:   ↔ wb 176  [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:  sbrbif  2074  sbabel  2515
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