Theorem sbss 4499

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		sseq1 3989	. 2 ⊢ (𝑥 = 𝑧 → (𝑥 ⊆ 𝐴 ↔ 𝑧 ⊆ 𝐴))
2		sseq1 3989	. 2 ⊢ (𝑧 = 𝑦 → (𝑧 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴))
3	1, 2	sbievw2 2099	1 ⊢ ([𝑦 / 𝑥]𝑥 ⊆ 𝐴 ↔ 𝑦 ⊆ 𝐴)

Syntax hints:   ↔ wb 206  [wsb 2065   ⊆ wss 3931

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-9 2119  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-sb 2066  df-cleq 2728  df-ss 3948

This theorem is referenced by: (None)