Theorem sbal2 1939

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		alcom 1407	. . 3 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
2		hbnae 1649	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
3		dveeq1 1936	. . . . . . 7 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
4	3	alimi 1384	. . . . . 6 ⊢ (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
5	4	hbnaes 1651	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
6		19.21ht 1513	. . . . 5 ⊢ (∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧) → (∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
7	5, 6	syl 14	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ (𝑦 = 𝑧 → ∀𝑥𝜑)))
8	2, 7	albidh 1409	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
9	1, 8	syl5rbbr 193	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑)))
10		sb6 1807	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11		sb6 1807	. . 3 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → 𝜑))
12	11	albii 1399	. 2 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
13	9, 10, 12	3bitr4g 221	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 103  ∀wal 1282  [wsb 1685

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-fal 1290  df-nf 1390  df-sb 1686

This theorem is referenced by: (None)