Theorem sbalyz 1975

Description: Move universal quantifier in and out of substitution. Identical to sbal 1976 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 1522	. . . 4 ⊢ Ⅎ𝑥∀𝑥𝜑
2	1	nfsbxy 1916	. . 3 ⊢ Ⅎ𝑥[𝑧 / 𝑦]∀𝑥𝜑
3		ax-4 1488	. . . 4 ⊢ (∀𝑥𝜑 → 𝜑)
4	3	sbimi 1738	. . 3 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 → [𝑧 / 𝑦]𝜑)
5	2, 4	alrimi 1503	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 → ∀𝑥[𝑧 / 𝑦]𝜑)
6		sb6 1859	. . . . 5 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → 𝜑))
7	6	albii 1447	. . . 4 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
8		alcom 1455	. . . 4 ⊢ (∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑))
9	7, 8	bitri 183	. . 3 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑))
10		nfv 1509	. . . . . 6 ⊢ Ⅎ𝑥 𝑦 = 𝑧
11	10	stdpc5 1564	. . . . 5 ⊢ (∀𝑥(𝑦 = 𝑧 → 𝜑) → (𝑦 = 𝑧 → ∀𝑥𝜑))
12	11	alimi 1432	. . . 4 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) → ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
13		sb2 1741	. . . 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 120	. 2 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 → [𝑧 / 𝑦]∀𝑥𝜑)
16	5, 15	impbii 125	1 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑)

Syntax hints:   → wi 4   ↔ wb 104  ∀wal 1330  [wsb 1736

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-bndl 1487  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516

This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737

This theorem is referenced by:  sbal  1976