Theorem sbss 3357

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 2605	. 2 ⊢ 𝑦 ∈ V
2		sbequ 1762	. 2 ⊢ (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥 ⊆ 𝐴 ↔ [𝑦 / 𝑥]𝑥 ⊆ 𝐴))
3		sseq1 3021	. 2 ⊢ (𝑧 = 𝑦 → (𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴))
4		nfv 1462	. . 3 ⊢ Ⅎ𝑥 𝑧 ⊆ 𝐴
5		sseq1 3021	. . 3 ⊢ (𝑥 = 𝑧 → (𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴))
6	4, 5	sbie 1715	. 2 ⊢ ([𝑧 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴)
7	1, 2, 3, 6	vtoclb 2657	1 ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)

Syntax hints:   ↔ wb 103  [wsb 1686   ⊆ wss 2974

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 663  ax-5 1377  ax-7 1378  ax-gen 1379  ax-ie1 1423  ax-ie2 1424  ax-8 1436  ax-10 1437  ax-11 1438  ax-i12 1439  ax-4 1441  ax-17 1460  ax-i9 1464  ax-ial 1468  ax-i5r 1469  ax-ext 2064

This theorem depends on definitions:  df-bi 115  df-nf 1391  df-sb 1687  df-clab 2069  df-cleq 2075  df-clel 2078  df-v 2604  df-in 2980  df-ss 2987

This theorem is referenced by: (None)