Theorem sbcgf 1469

Description: Substitution for a variable not free in a wff does not affect it.

Hypothesis

Ref	Expression
sbcgf.1	⊢ (φ → ∀xφ)

Assertion

Ref	Expression
sbcgf	⊢ (A ∈ B → ([A / x]φ ↔ φ))

Proof of Theorem sbcgf

Step	Hyp	Ref	Expression
1		sbcco 1448	. 2 ⊢ (A ∈ B → ([A / y][y / x]φ ↔ [A / x]φ))
2		eqid 810	. . . 4 ⊢ x = x
3		sbcgf.1	. . . . . . 7 ⊢ (φ → ∀xφ)
4	3	sbf 870	. . . . . 6 ⊢ ([y / x]φ ↔ φ)
5	4	a1i 7	. . . . 5 ⊢ (x = x → ([y / x]φ ↔ φ))
6	5	bisbcdv 1468	. . . 4 ⊢ ((A ∈ B ∧ x = x) → ([A / y][y / x]φ ↔ [A / y]φ))
7	2, 6	mpan2 519	. . 3 ⊢ (A ∈ B → ([A / y][y / x]φ ↔ [A / y]φ))
8		sbc5g 1450	. . 3 ⊢ (A ∈ B → ([A / y]φ ↔ ∃y(y = A ∧ φ)))
9		elex 1356	. . . . 5 ⊢ (A ∈ B → ∃y y = A)
10	9	biantrurd 546	. . . 4 ⊢ (A ∈ B → (φ ↔ (∃y y = A ∧ φ)))
11		19.41v 963	. . . 4 ⊢ (∃y(y = A ∧ φ) ↔ (∃y y = A ∧ φ))
12	10, 11	syl6rbbr 417	. . 3 ⊢ (A ∈ B → (∃y(y = A ∧ φ) ↔ φ))
13	7, 8, 12	3bitrd 422	. 2 ⊢ (A ∈ B → ([A / y][y / x]φ ↔ φ))
14	1, 13	bitr3d 408	1 ⊢ (A ∈ B → ([A / x]φ ↔ φ))

Syntax hints:   → wi 2   ↔ wb 127   ∧ wa 196  ∀wal 672  ∃wex 678   = weq 797  [wsb 852   = wceq 1091   ∈ wcel 1092  [wsbc 1440

This theorem is referenced by:  sbc19.21g 1470

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-v 1349  df-sbc 1441

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