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Theorem bj-sbnf 33260
Description: Move non-free predicate in and out of substitution; see sbal 2554 and sbex 2555. (Contributed by BJ, 2-May-2019.)
Assertion
Ref Expression
bj-sbnf ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ Ⅎ𝑥[𝑧 / 𝑦]𝜑)
Distinct variable groups:   𝑥,𝑦   𝑥,𝑧
Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)

Proof of Theorem bj-sbnf
StepHypRef Expression
1 sbim 2486 . . . 4 ([𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑))
2 sbal 2554 . . . . 5 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)
32imbi2i 327 . . . 4 (([𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
41, 3bitri 266 . . 3 ([𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
54albii 1914 . 2 (∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ∀𝑥([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
6 nf5 2292 . . . 4 (Ⅎ𝑥𝜑 ↔ ∀𝑥(𝜑 → ∀𝑥𝜑))
76sbbii 2069 . . 3 ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ [𝑧 / 𝑦]∀𝑥(𝜑 → ∀𝑥𝜑))
8 sbal 2554 . . 3 ([𝑧 / 𝑦]∀𝑥(𝜑 → ∀𝑥𝜑) ↔ ∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑))
97, 8bitri 266 . 2 ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑))
10 nf5 2292 . 2 (Ⅎ𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
115, 9, 103bitr4i 294 1 ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ Ⅎ𝑥[𝑧 / 𝑦]𝜑)
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 197  wal 1650  wnf 1878  [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:  bj-nfcf  33350
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