Theorem sbss 3660

Description: Set substitution into the first argument of a subset relation. (Contributed by Rodolfo Medina, 7-Jul-2010.) (Proof shortened by Mario Carneiro, 14-Nov-2016.)

Assertion

Ref	Expression
sbss	⊢ ([y / x]x ⊆ A ↔ y ⊆ A)

Distinct variable group:   x,A

Allowed substitution hint:   A(y)

Proof of Theorem sbss

Dummy variable z is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 2863	. 2 ⊢ y ∈ V
2		sbequ 2060	. 2 ⊢ (z = y → ([z / x]x ⊆ A ↔ [y / x]x ⊆ A))
3		sseq1 3293	. 2 ⊢ (z = y → (z ⊆ A ↔ y ⊆ A))
4		nfv 1619	. . 3 ⊢ Ⅎx z ⊆ A
5		sseq1 3293	. . 3 ⊢ (x = z → (x ⊆ A ↔ z ⊆ A))
6	4, 5	sbie 2038	. 2 ⊢ ([z / x]x ⊆ A ↔ z ⊆ A)
7	1, 2, 3, 6	vtoclb 2913	1 ⊢ ([y / x]x ⊆ A ↔ y ⊆ A)

Syntax hints:   ↔ wb 176  [wsb 1648   ⊆ wss 3258

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-ss 3260

This theorem is referenced by: (None)