Theorem sb6y 872

Description: Equivalence involving substitution with a variable not free.

Hypothesis

Ref	Expression
sb6y.1	⊢ (φ → ∀yφ)

Assertion

Ref	Expression
sb6y	⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Proof of Theorem sb6y

Step	Hyp	Ref	Expression
1		sbequ2 864	. . . . 5 ⊢ (x = y → ([y / x]φ → φ))
2	1	a4s 682	. . . 4 ⊢ (∀x x = y → ([y / x]φ → φ))
3		ax-10 800	. . . . . . 7 ⊢ (∀y y = x → (∀yφ → ∀xφ))
4	3	eq4s 822	. . . . . 6 ⊢ (∀x x = y → (∀yφ → ∀xφ))
5		ax-1 3	. . . . . . 7 ⊢ (φ → (x = y → φ))
6	5	19.20i 691	. . . . . 6 ⊢ (∀xφ → ∀x(x = y → φ))
7	4, 6	syl6 23	. . . . 5 ⊢ (∀x x = y → (∀yφ → ∀x(x = y → φ)))
8		sb6y.1	. . . . 5 ⊢ (φ → ∀yφ)
9	7, 8	syl5 22	. . . 4 ⊢ (∀x x = y → (φ → ∀x(x = y → φ)))
10	2, 9	syld 27	. . 3 ⊢ (∀x x = y → ([y / x]φ → ∀x(x = y → φ)))
11		sb4 861	. . 3 ⊢ (¬ ∀x x = y → ([y / x]φ → ∀x(x = y → φ)))
12	10, 11	pm2.61i 110	. 2 ⊢ ([y / x]φ → ∀x(x = y → φ))
13		sb2 859	. 2 ⊢ (∀x(x = y → φ) → [y / x]φ)
14	12, 13	impbi 139	1 ⊢ ([y / x]φ ↔ ∀x(x = y → φ))

Syntax hints:   → wi 2   ↔ wb 127  ∀wal 672   = weq 797  [wsb 852

This theorem is referenced by:  sb5y 883

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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