Theorem sbn2 881

Description: Introduction of negation into substitution.

Assertion

Ref	Expression
sbn2	⊢ (¬ [y / x]φ → [y / x] ¬ φ)

Proof of Theorem sbn2

Step	Hyp	Ref	Expression
1		sbequ1 863	. . . . 5 ⊢ (x = y → (φ → [y / x]φ))
2	1	con3d 87	. . . 4 ⊢ (x = y → (¬ [y / x]φ → ¬ φ))
3	2	com12 13	. . 3 ⊢ (¬ [y / x]φ → (x = y → ¬ φ))
4		sb2 859	. . . . . 6 ⊢ (∀x(x = y → ¬ ¬ φ) → [y / x] ¬ ¬ φ)
5		pm4.13 142	. . . . . . 7 ⊢ (φ ↔ ¬ ¬ φ)
6	5	bisb 855	. . . . . 6 ⊢ ([y / x]φ ↔ [y / x] ¬ ¬ φ)
7	4, 6	sylibr 175	. . . . 5 ⊢ (∀x(x = y → ¬ ¬ φ) → [y / x]φ)
8	7	con3i 90	. . . 4 ⊢ (¬ [y / x]φ → ¬ ∀x(x = y → ¬ ¬ φ))
9		eqs3 830	. . . 4 ⊢ (∃x(x = y ∧ ¬ φ) ↔ ¬ ∀x(x = y → ¬ ¬ φ))
10	8, 9	sylibr 175	. . 3 ⊢ (¬ [y / x]φ → ∃x(x = y ∧ ¬ φ))
11	3, 10	jca 236	. 2 ⊢ (¬ [y / x]φ → ((x = y → ¬ φ) ∧ ∃x(x = y ∧ ¬ φ)))
12		df-sb 853	. 2 ⊢ ([y / x] ¬ φ ↔ ((x = y → ¬ φ) ∧ ∃x(x = y ∧ ¬ φ)))
13	11, 12	sylibr 175	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-gen 677  ax-9 799

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

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