Theorem hbsb3 875

Description: If y is not free in φ, x is not free in [y / x]φ.

Hypothesis

Ref	Expression
hbsb3.1	⊢ (φ → ∀yφ)

Assertion

Ref	Expression
hbsb3	⊢ ([y / x]φ → ∀x[y / x]φ)

Proof of Theorem hbsb3

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	. . . . . 6 ⊢ (∀y y = x → (∀yφ → ∀xφ))
4	3	eq4s 822	. . . . 5 ⊢ (∀x x = y → (∀yφ → ∀xφ))
5		hbsb3.1	. . . . 5 ⊢ (φ → ∀yφ)
6	4, 5	syl5 22	. . . 4 ⊢ (∀x x = y → (φ → ∀xφ))
7	2, 6	syld 27	. . 3 ⊢ (∀x x = y → ([y / x]φ → ∀xφ))
8		sbequ1 863	. . . 4 ⊢ (x = y → (φ → [y / x]φ))
9	8	19.20ii 692	. . 3 ⊢ (∀x x = y → (∀xφ → ∀x[y / x]φ))
10	7, 9	syld 27	. 2 ⊢ (∀x x = y → ([y / x]φ → ∀x[y / x]φ))
11		hbsb2 873	. 2 ⊢ (¬ ∀x x = y → ([y / x]φ → ∀x[y / x]φ))
12	10, 11	pm2.61i 110	1 ⊢ ([y / x]φ → ∀x[y / x]φ)

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

This theorem is referenced by:  sbco2 913  sb8 918  mo 1020  axrepndlem1 3738  axrepndlem2 3739

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

metamath.org