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Theorem bj-sbnf 34067
 Description: Move nonfree predicate in and out of substitution; see sbal 2158 and sbex 2282. (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 2305 . . . 4 ([𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑))
2 sbal 2158 . . . . 5 ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)
32imbi2i 337 . . . 4 (([𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
41, 3bitri 276 . . 3 ([𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
54albii 1813 . 2 (∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ∀𝑥([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
6 nf5 2284 . . . 4 (Ⅎ𝑥𝜑 ↔ ∀𝑥(𝜑 → ∀𝑥𝜑))
76sbbii 2074 . . 3 ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ [𝑧 / 𝑦]∀𝑥(𝜑 → ∀𝑥𝜑))
8 sbal 2158 . . 3 ([𝑧 / 𝑦]∀𝑥(𝜑 → ∀𝑥𝜑) ↔ ∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑))
97, 8bitri 276 . 2 ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑))
10 nf5 2284 . 2 (Ⅎ𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
115, 9, 103bitr4i 304 1 ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ Ⅎ𝑥[𝑧 / 𝑦]𝜑)
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ↔ wb 207  ∀wal 1528  Ⅎwnf 1777  [wsb 2062 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1904  ax-6 1963  ax-7 2008  ax-10 2138  ax-11 2153  ax-12 2169 This theorem depends on definitions:  df-bi 208  df-an 397  df-or 844  df-ex 1774  df-nf 1778  df-sb 2063 This theorem is referenced by:  bj-nfcf  34145
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