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Theorem sbhb 2556
Description: Two ways of expressing "𝑥 is (effectively) not free in 𝜑". (Contributed by NM, 29-May-2009.)
Assertion
Ref Expression
sbhb ((𝜑 → ∀𝑥𝜑) ↔ ∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑))
Distinct variable group:   𝜑,𝑦
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem sbhb
StepHypRef Expression
1 nfv 2010 . . . 4 𝑦𝜑
21sb8 2535 . . 3 (∀𝑥𝜑 ↔ ∀𝑦[𝑦 / 𝑥]𝜑)
32imbi2i 328 . 2 ((𝜑 → ∀𝑥𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
4 19.21v 2035 . 2 (∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
53, 4bitr4i 270 1 ((𝜑 → ∀𝑥𝜑) ↔ ∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 198  wal 1651  [wsb 2064
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-10 2185  ax-11 2200  ax-12 2213  ax-13 2354
This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-ex 1876  df-nf 1880  df-sb 2065
This theorem is referenced by: (None)
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