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Theorem sbbid 2402
Description: Deduction substituting both sides of a biconditional. (Contributed by NM, 30-Jun-1993.)
Hypotheses
Ref Expression
sbbid.1 𝑥𝜑
sbbid.2 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
sbbid (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))

Proof of Theorem sbbid
StepHypRef Expression
1 sbbid.1 . . 3 𝑥𝜑
2 sbbid.2 . . 3 (𝜑 → (𝜓𝜒))
31, 2alrimi 2080 . 2 (𝜑 → ∀𝑥(𝜓𝜒))
4 spsbbi 2401 . 2 (∀𝑥(𝜓𝜒) → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
53, 4syl 17 1 (𝜑 → ([𝑦 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜒))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 196  wal 1478  wnf 1705  [wsb 1877
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1719  ax-4 1734  ax-5 1836  ax-6 1885  ax-7 1932  ax-10 2016  ax-12 2044  ax-13 2245
This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-ex 1702  df-nf 1707  df-sb 1878
This theorem is referenced by:  sbcom3  2410  sbco3  2416  sbcom2  2444  sbal  2461  wl-equsb3  33004  wl-sbcom2d-lem1  33009  wl-sbcom3  33039
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