Theorem sbcssg 3617

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 3094	. . 3 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦[𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶)))
2		sbcimg 3083	. . . . 5 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ([𝐴 / 𝑥]𝑦 ∈ 𝐵 → [𝐴 / 𝑥]𝑦 ∈ 𝐶)))
3		sbcel2g 3158	. . . . . 6 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝑦 ∈ 𝐵 ↔ 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵))
4		sbcel2g 3158	. . . . . 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 1873	. . 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 3225	. . 3 ⊢ (𝐵 ⊆ 𝐶 ↔ ∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
10	9	sbcbii 3101	. 2 ⊢ ([𝐴 / 𝑥]𝐵 ⊆ 𝐶 ↔ [𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
11		ssalel 3225	. 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 1396   ∈ wcel 2203  [wsbc 3041  ⦋csb 3137   ⊆ wss 3210

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2814  df-sbc 3042  df-csb 3138  df-in 3216  df-ss 3223

This theorem is referenced by:  sbcrel  4835  sbcfg  5506