Theorem sbss 3349

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 2604	. 2 ⊢ 𝑦 ∈ V
2		sbequ 1761	. 2 ⊢ (𝑧 = 𝑦 → ([𝑧 / 𝑥]𝑥 ⊆ 𝐴 ↔ [𝑦 / 𝑥]𝑥 ⊆ 𝐴))
3		sseq1 3020	. 2 ⊢ (𝑧 = 𝑦 → (𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴))
4		nfv 1461	. . 3 ⊢ Ⅎ𝑥 𝑧 ⊆ 𝐴
5		sseq1 3020	. . 3 ⊢ (𝑥 = 𝑧 → (𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴))
6	4, 5	sbie 1714	. 2 ⊢ ([𝑧 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴)
7	1, 2, 3, 6	vtoclb 2656	1 ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)

Syntax hints:   ↔ wb 103  [wsb 1685   ⊆ wss 2973

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 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-v 2603  df-in 2979  df-ss 2986

This theorem is referenced by: (None)