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Theorem sbhb 1859
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 ax-17 1460 . . . 4 (𝜑 → ∀𝑦𝜑)
21sb8h 1777 . . 3 (∀𝑥𝜑 ↔ ∀𝑦[𝑦 / 𝑥]𝜑)
32imbi2i 224 . 2 ((𝜑 → ∀𝑥𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
4 19.21v 1796 . 2 (∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑) ↔ (𝜑 → ∀𝑦[𝑦 / 𝑥]𝜑))
53, 4bitr4i 185 1 ((𝜑 → ∀𝑥𝜑) ↔ ∀𝑦(𝜑 → [𝑦 / 𝑥]𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wb 103  wal 1283  [wsb 1687
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-11 1438  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469
This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1688
This theorem is referenced by:  sbnf2  1900
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