MPE Home Metamath Proof Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  MPE Home  >  Th. List  >  sbrbis Structured version   Visualization version   GIF version

Theorem sbrbis 2392
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 2388 . 2 ([𝑦 / 𝑥](𝜑𝜒) ↔ ([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒))
2 sbrbis.1 . . 3 ([𝑦 / 𝑥]𝜑𝜓)
32bibi1i 326 . 2 (([𝑦 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
41, 3bitri 262 1 ([𝑦 / 𝑥](𝜑𝜒) ↔ (𝜓 ↔ [𝑦 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wb 194  [wsb 1866
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1712  ax-4 1727  ax-5 1826  ax-6 1874  ax-7 1921  ax-10 2005  ax-12 2033  ax-13 2233
This theorem depends on definitions:  df-bi 195  df-or 383  df-an 384  df-tru 1477  df-ex 1695  df-nf 1700  df-sb 1867
This theorem is referenced by:  sbrbif  2393  sbabel  2778
  Copyright terms: Public domain W3C validator