Theorem sbn1 880

Description: Removal of negation from substitution.

Assertion

Ref	Expression
sbn1	⊢ ([y / x] ¬ φ → ¬ [y / x]φ)

Proof of Theorem sbn1

Step	Hyp	Ref	Expression
1		sbequ2 864	. . . 4 ⊢ (x = y → ([y / x] ¬ φ → ¬ φ))
2		sbequ2 864	. . . . 5 ⊢ (x = y → ([y / x]φ → φ))
3	2	con3d 87	. . . 4 ⊢ (x = y → (¬ φ → ¬ [y / x]φ))
4	1, 3	syld 27	. . 3 ⊢ (x = y → ([y / x] ¬ φ → ¬ [y / x]φ))
5	4	a4s 682	. 2 ⊢ (∀x x = y → ([y / x] ¬ φ → ¬ [y / x]φ))
6		sb4 861	. . 3 ⊢ (¬ ∀x x = y → ([y / x] ¬ φ → ∀x(x = y → ¬ φ)))
7		sb1 858	. . . . 5 ⊢ ([y / x]φ → ∃x(x = y ∧ φ))
8		eqs3 830	. . . . 5 ⊢ (∃x(x = y ∧ φ) ↔ ¬ ∀x(x = y → ¬ φ))
9	7, 8	sylib 173	. . . 4 ⊢ ([y / x]φ → ¬ ∀x(x = y → ¬ φ))
10	9	con2i 89	. . 3 ⊢ (∀x(x = y → ¬ φ) → ¬ [y / x]φ)
11	6, 10	syl6 23	. 2 ⊢ (¬ ∀x x = y → ([y / x] ¬ φ → ¬ [y / x]φ))
12	5, 11	pm2.61i 110	1 ⊢ ([y / x] ¬ φ → ¬ [y / x]φ)

Syntax hints:  ¬ wn 1   → wi 2   ∧ wa 196  ∀wal 672  ∃wex 678   = weq 797  [wsb 852

This theorem is referenced by:  sbn 882

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

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853

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