Theorem sbal2 2013

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 1714	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
2		dveeq1 2012	. . . . . . 7 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
3	2	alimi 1448	. . . . . 6 ⊢ (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
4	3	hbnaes 1716	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
5		19.21ht 1574	. . . . 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 1473	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
8		alcom 1471	. . 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 1879	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11		sb6 1879	. . 3 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → 𝜑))
12	11	albii 1463	. 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 1346  [wsb 1755

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 609  ax-in2 610  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528

This theorem depends on definitions:  df-bi 116  df-tru 1351  df-fal 1354  df-nf 1454  df-sb 1756

This theorem is referenced by: (None)