Theorem spc3gv 3558

Description: Specialization with three quantifiers, using implicit substitution. (Contributed by NM, 12-May-2008.)

Hypothesis

Ref	Expression
spc3egv.1	⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝑧 = 𝐶) → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
spc3gv	⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → (∀𝑥∀𝑦∀𝑧𝜑 → 𝜓))

Distinct variable groups:   𝑥,𝑦,𝑧,𝐴   𝑥,𝐵,𝑦,𝑧   𝑥,𝐶,𝑦,𝑧   𝜓,𝑥,𝑦,𝑧

Allowed substitution hints:   𝜑(𝑥, 𝑦, 𝑧)   𝑉(𝑥, 𝑦, 𝑧)   𝑊(𝑥, 𝑦, 𝑧)   𝑋(𝑥, 𝑦, 𝑧)

Proof of Theorem spc3gv

Step	Hyp	Ref	Expression
1		spc3egv.1	. . . . 5 ⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝑧 = 𝐶) → (𝜑 ↔ 𝜓))
2	1	notbid 321	. . . 4 ⊢ ((𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ∧ 𝑧 = 𝐶) → (¬ 𝜑 ↔ ¬ 𝜓))
3	2	spc3egv 3557	. . 3 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → (¬ 𝜓 → ∃𝑥∃𝑦∃𝑧 ¬ 𝜑))
4		exnal 1860	. . . . . . 7 ⊢ (∃𝑧 ¬ 𝜑 ↔ ¬ ∀𝑧𝜑)
5	4	exbii 1881	. . . . . 6 ⊢ (∃𝑦∃𝑧 ¬ 𝜑 ↔ ∃𝑦 ¬ ∀𝑧𝜑)
6		exnal 1860	. . . . . 6 ⊢ (∃𝑦 ¬ ∀𝑧𝜑 ↔ ¬ ∀𝑦∀𝑧𝜑)
7	5, 6	bitri 278	. . . . 5 ⊢ (∃𝑦∃𝑧 ¬ 𝜑 ↔ ¬ ∀𝑦∀𝑧𝜑)
8	7	exbii 1881	. . . 4 ⊢ (∃𝑥∃𝑦∃𝑧 ¬ 𝜑 ↔ ∃𝑥 ¬ ∀𝑦∀𝑧𝜑)
9		exnal 1860	. . . 4 ⊢ (∃𝑥 ¬ ∀𝑦∀𝑧𝜑 ↔ ¬ ∀𝑥∀𝑦∀𝑧𝜑)
10	8, 9	bitr2i 279	. . 3 ⊢ (¬ ∀𝑥∀𝑦∀𝑧𝜑 ↔ ∃𝑥∃𝑦∃𝑧 ¬ 𝜑)
11	3, 10	imbitrrdi 255	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → (¬ 𝜓 → ¬ ∀𝑥∀𝑦∀𝑧𝜑))
12	11	con4d 116	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝐶 ∈ 𝑋) → (∀𝑥∀𝑦∀𝑧𝜑 → 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ w3a 1103  ∀wal 1568   = wceq 1570  ∃wex 1812   ∈ wcel 2145

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-3an 1105  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452

This theorem is used by:  funopg  6562  pslem  18708  dirtr  18738  mclsax  36255  fununiq  36455