Theorem sbalyz 1999

Description: Move universal quantifier in and out of substitution. Identical to sbal 2000 except that it has an additional distinct variable constraint on 𝑦 and 𝑧. (Contributed by Jim Kingdon, 29-Dec-2017.)

Assertion

Ref	Expression
sbalyz	⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)

Distinct variable group:   𝑥,𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥,𝑦,𝑧)

Proof of Theorem sbalyz

Step	Hyp	Ref	Expression
1		nfa1 1541	. . . 4 ⊢ Ⅎ𝑥∀𝑥𝜑
2	1	nfsbxy 1942	. . 3 ⊢ Ⅎ𝑥[𝑧 / 𝑦]∀𝑥𝜑
3		ax-4 1510	. . . 4 ⊢ (∀𝑥𝜑 → 𝜑)
4	3	sbimi 1764	. . 3 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 → [𝑧 / 𝑦]𝜑)
5	2, 4	alrimi 1522	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑)
6		sb6 1886	. . . . 5 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → 𝜑))
7	6	albii 1470	. . . 4 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
8		alcom 1478	. . . 4 ⊢ (∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑))
9	7, 8	bitri 184	. . 3 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑))
10		nfv 1528	. . . . . 6 ⊢ Ⅎ𝑥 𝑦 = 𝑧
11	10	stdpc5 1584	. . . . 5 ⊢ (∀𝑥(𝑦 = 𝑧 → 𝜑) → (𝑦 = 𝑧 → ∀𝑥𝜑))
12	11	alimi 1455	. . . 4 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) → ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
13		sb2 1767	. . . 4 ⊢ (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) → [𝑧 / 𝑦]∀𝑥𝜑)
14	12, 13	syl 14	. . 3 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) → [𝑧 / 𝑦]∀𝑥𝜑)
15	9, 14	sylbi 121	. 2 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑)
16	5, 15	impbii 126	1 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1351  [wsb 1762

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535

This theorem depends on definitions:  df-bi 117  df-nf 1461  df-sb 1763

This theorem is referenced by:  sbal  2000