Theorem sbal2 2008

Description: Move quantifier in and out of substitution. (Contributed by NM, 2-Jan-2002.)

Assertion

Ref	Expression
sbal2	⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))

Distinct variable groups:   𝑦,𝑧   𝑥,𝑧

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

Proof of Theorem sbal2

Step	Hyp	Ref	Expression
1		hbnae 1709	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
2		dveeq1 2007	. . . . . . 7 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
3	2	alimi 1443	. . . . . 6 ⊢ (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
4	3	hbnaes 1711	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
5		19.21ht 1569	. . . . 5 ⊢ (∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧) → (∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
6	4, 5	syl 14	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
7	1, 6	albidh 1468	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
8		alcom 1466	. . 3 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
9	7, 8	bitr3di 194	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑)))
10		sb6 1874	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11		sb6 1874	. . 3 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → 𝜑))
12	11	albii 1458	. 2 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
13	9, 10, 12	3bitr4g 222	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 104  ∀wal 1341  [wsb 1750

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-in1 604  ax-in2 605  ax-io 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523

This theorem depends on definitions:  df-bi 116  df-tru 1346  df-fal 1349  df-nf 1449  df-sb 1751

This theorem is referenced by: (None)