Theorem bj-sbnf 32828

Description: Move non-free predicate in and out of substitution; see sbal 2462 and sbex 2463. (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

Step	Hyp	Ref	Expression
1		sbim 2395	. . . 4 ⊢ ([𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑))
2		sbal 2462	. . . . 5 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)
3	2	imbi2i 326	. . . 4 ⊢ (([𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
4	1, 3	bitri 264	. . 3 ⊢ ([𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
5	4	albii 1747	. 2 ⊢ (∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑) ↔ ∀𝑥([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
6		nf5 2116	. . . 4 ⊢ (Ⅎ𝑥𝜑 ↔ ∀𝑥(𝜑 → ∀𝑥𝜑))
7	6	sbbii 1887	. . 3 ⊢ ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ [𝑧 / 𝑦]∀𝑥(𝜑 → ∀𝑥𝜑))
8		sbal 2462	. . 3 ⊢ ([𝑧 / 𝑦]∀𝑥(𝜑 → ∀𝑥𝜑) ↔ ∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑))
9	7, 8	bitri 264	. 2 ⊢ ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦](𝜑 → ∀𝑥𝜑))
10		nf5 2116	. 2 ⊢ (Ⅎ𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥([𝑧 / 𝑦]𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑))
11	5, 9, 10	3bitr4i 292	1 ⊢ ([𝑧 / 𝑦]Ⅎ𝑥𝜑 ↔ Ⅎ𝑥[𝑧 / 𝑦]𝜑)

Syntax hints:   → wi 4   ↔ wb 196  ∀wal 1481  Ⅎwnf 1708  [wsb 1880

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881

This theorem is referenced by:  bj-nfcf  32920