Theorem sbss 3599

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 2802	. 2 ⊢ 𝑦 ∈ V
2		sbequ 1886	. 2 ⊢ (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥 ⊆ 𝐴 ↔ [𝑦 / 𝑥]𝑥 ⊆ 𝐴))
3		sseq1 3247	. 2 ⊢ (𝑧 = 𝑦 → (𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴))
4		nfv 1574	. . 3 ⊢ Ⅎ𝑥 𝑧 ⊆ 𝐴
5		sseq1 3247	. . 3 ⊢ (𝑥 = 𝑧 → (𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴))
6	4, 5	sbie 1837	. 2 ⊢ ([𝑧 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴)
7	1, 2, 3, 6	vtoclb 2858	1 ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)

Syntax hints:   ↔ wb 105  [wsb 1808   ⊆ wss 3197

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-v 2801  df-in 3203  df-ss 3210

This theorem is referenced by: (None)