Theorem spc3gv 2945

Description: Specialization with 3 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 285	. . . 4 ⊢ ((x = A ∧ y = B ∧ z = C) → (¬ φ ↔ ¬ ψ))
3	2	spc3egv 2944	. . 3 ⊢ ((A ∈ V ∧ B ∈ W ∧ C ∈ X) → (¬ ψ → ∃x∃y∃z ¬ φ))
4		exnal 1574	. . . . . . 7 ⊢ (∃z ¬ φ ↔ ¬ ∀zφ)
5	4	exbii 1582	. . . . . 6 ⊢ (∃y∃z ¬ φ ↔ ∃y ¬ ∀zφ)
6		exnal 1574	. . . . . 6 ⊢ (∃y ¬ ∀zφ ↔ ¬ ∀y∀zφ)
7	5, 6	bitri 240	. . . . 5 ⊢ (∃y∃z ¬ φ ↔ ¬ ∀y∀zφ)
8	7	exbii 1582	. . . 4 ⊢ (∃x∃y∃z ¬ φ ↔ ∃x ¬ ∀y∀zφ)
9		exnal 1574	. . . 4 ⊢ (∃x ¬ ∀y∀zφ ↔ ¬ ∀x∀y∀zφ)
10	8, 9	bitr2i 241	. . 3 ⊢ (¬ ∀x∀y∀zφ ↔ ∃x∃y∃z ¬ φ)
11	3, 10	syl6ibr 218	. 2 ⊢ ((A ∈ V ∧ B ∈ W ∧ C ∈ X) → (¬ ψ → ¬ ∀x∀y∀zφ))
12	11	con4d 97	1 ⊢ ((A ∈ V ∧ B ∈ W ∧ C ∈ X) → (∀x∀y∀zφ → ψ))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∧ w3a 934  ∀wal 1540  ∃wex 1541   = wceq 1642   ∈ wcel 1710

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-ext 2334

This proof depends on definitions:  df-bi 177  df-an 360  df-3an 936  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-v 2862

This theorem is used by:  fununiq  5518