Theorem sbcssg 3580

Description: Distribute proper substitution through a subclass relation. (Contributed by Alan Sare, 22-Jul-2012.) (Proof shortened by Alexander van der Vekens, 23-Jul-2017.)

Assertion

Ref	Expression
sbcssg	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ⊆ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ⊆ ⦋𝐴 / 𝑥⦌𝐶))

Proof of Theorem sbcssg

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcalg 3061	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦[𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶)))
2		sbcimg 3050	. . . . 5 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ([𝐴 / 𝑥]𝑦 ∈ 𝐵 → [𝐴 / 𝑥]𝑦 ∈ 𝐶)))
3		sbcel2g 3125	. . . . . 6 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑦 ∈ 𝐵 ↔ 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵))
4		sbcel2g 3125	. . . . . 6 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑦 ∈ 𝐶 ↔ 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶))
5	3, 4	imbi12d 234	. . . . 5 ⊢ (𝐴 ∈ 𝑉 → (([𝐴 / 𝑥]𝑦 ∈ 𝐵 → [𝐴 / 𝑥]𝑦 ∈ 𝐶) ↔ (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
6	2, 5	bitrd 188	. . . 4 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
7	6	albidv 1850	. . 3 ⊢ (𝐴 ∈ 𝑉 → (∀𝑦[𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦(𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
8	1, 7	bitrd 188	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦(𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
9		ssalel 3192	. . 3 ⊢ (𝐵 ⊆ 𝐶 ↔ ∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
10	9	sbcbii 3068	. 2 ⊢ ([𝐴 / 𝑥]𝐵 ⊆ 𝐶 ↔ [𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
11		ssalel 3192	. 2 ⊢ (⦋𝐴 / 𝑥⦌𝐵 ⊆ ⦋𝐴 / 𝑥⦌𝐶 ↔ ∀𝑦(𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶))
12	8, 10, 11	3bitr4g 223	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ⊆ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ⊆ ⦋𝐴 / 𝑥⦌𝐶))

Syntax hints:   → wi 4   ↔ wb 105  ∀wal 1373   ∈ wcel 2180  [wsbc 3008  ⦋csb 3104   ⊆ wss 3177

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 713  ax-5 1473  ax-7 1474  ax-gen 1475  ax-ie1 1519  ax-ie2 1520  ax-8 1530  ax-10 1531  ax-11 1532  ax-i12 1533  ax-bndl 1535  ax-4 1536  ax-17 1552  ax-i9 1556  ax-ial 1560  ax-i5r 1561  ax-ext 2191

This theorem depends on definitions:  df-bi 117  df-tru 1378  df-nf 1487  df-sb 1789  df-clab 2196  df-cleq 2202  df-clel 2205  df-nfc 2341  df-v 2781  df-sbc 3009  df-csb 3105  df-in 3183  df-ss 3190

This theorem is referenced by:  sbcrel  4782  sbcfg  5448