Theorem sbal2 2080

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 1773	. . . 4 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑦 ¬ ∀𝑥 𝑥 = 𝑦)
2		dveeq1 2079	. . . . . . 7 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
3	2	alimi 1508	. . . . . 6 ⊢ (∀𝑥 ¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
4	3	hbnaes 1775	. . . . 5 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ∀𝑥(𝑦 = 𝑧 → ∀𝑥 𝑦 = 𝑧))
5		19.21ht 1634	. . . . 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 1533	. . 3 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑)))
8		alcom 1531	. . 3 ⊢ (∀𝑦∀𝑥(𝑦 = 𝑧 → 𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
9	7, 8	bitr3di 195	. 2 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → (∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑) ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑)))
10		sb6 1941	. 2 ⊢ ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → ∀𝑥𝜑))
11		sb6 1941	. . 3 ⊢ ([𝑧 / 𝑦]𝜑 ↔ ∀𝑦(𝑦 = 𝑧 → 𝜑))
12	11	albii 1523	. 2 ⊢ (∀𝑥[𝑧 / 𝑦]𝜑 ↔ ∀𝑥∀𝑦(𝑦 = 𝑧 → 𝜑))
13	9, 10, 12	3bitr4g 223	1 ⊢ (¬ ∀𝑥 𝑥 = 𝑦 → ([𝑧 / 𝑦]∀𝑥𝜑 ↔ ∀𝑥[𝑧 / 𝑦]𝜑))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 105  ∀wal 1400  [wsb 1815

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588

This proof depends on definitions:  df-bi 117  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816

This theorem is used by: (None)