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Theorem sbal 2554
Description: Move universal quantifier in and out of substitution. (Contributed by NM, 16-May-1993.) (Proof shortened by Wolf Lammen, 29-Sep-2018.)
Assertion
Ref Expression
sbal ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)
Distinct variable groups:   𝑥,𝑦   𝑥,𝑧
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)

Proof of Theorem sbal
StepHypRef Expression
1 nfae 2412 . . . 4 𝑦𝑥 𝑥 = 𝑧
2 axc16gb 2314 . . . 4 (∀𝑥 𝑥 = 𝑧 → (𝜑 ↔ ∀𝑥𝜑))
31, 2sbbid 2276 . . 3 (∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑦]𝜑 ↔ [𝑧 / 𝑦]∀𝑥𝜑))
4 axc16gb 2314 . . 3 (∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑦]𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))
53, 4bitr3d 272 . 2 (∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))
6 sbal1 2552 . 2 (¬ ∀𝑥 𝑥 = 𝑧 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))
75, 6pm2.61i 176 1 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)
Colors of variables: wff setvar class
Syntax hints:  wb 197  wal 1650  [wsb 2062
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1890  ax-4 1904  ax-5 2005  ax-6 2070  ax-7 2105  ax-10 2183  ax-11 2198  ax-12 2211  ax-13 2352
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 874  df-tru 1656  df-ex 1875  df-nf 1879  df-sb 2063
This theorem is referenced by:  sbex  2555  sbalv  2556  sbcal  3648  ax11-pm2  33255  bj-sbnf  33260
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