Theorem sbcssg 4487

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		sbcal 3806	. . 3 ⊢ ([𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦[𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
2		sbcimg 3795	. . . . 5 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ([𝐴 / 𝑥]𝑦 ∈ 𝐵 → [𝐴 / 𝑥]𝑦 ∈ 𝐶)))
3		sbcel2 4386	. . . . . 6 ⊢ ([𝐴 / 𝑥]𝑦 ∈ 𝐵 ↔ 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵)
4		sbcel2 4386	. . . . . 6 ⊢ ([𝐴 / 𝑥]𝑦 ∈ 𝐶 ↔ 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)
5	3, 4	imbi12i 353	. . . . 5 ⊢ (([𝐴 / 𝑥]𝑦 ∈ 𝐵 → [𝐴 / 𝑥]𝑦 ∈ 𝐶) ↔ (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶))
6	2, 5	bitrdi 290	. . . 4 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ (𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
7	6	albidv 1953	. . 3 ⊢ (𝐴 ∈ 𝑉 → (∀𝑦[𝐴 / 𝑥](𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦(𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
8	1, 7	bitrid 286	. 2 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶) ↔ ∀𝑦(𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶)))
9		df-ss 3925	. . 3 ⊢ (𝐵 ⊆ 𝐶 ↔ ∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
10	9	sbcbii 3803	. 2 ⊢ ([𝐴 / 𝑥]𝐵 ⊆ 𝐶 ↔ [𝐴 / 𝑥]∀𝑦(𝑦 ∈ 𝐵 → 𝑦 ∈ 𝐶))
11		df-ss 3925	. 2 ⊢ (⦋𝐴 / 𝑥⦌𝐵 ⊆ ⦋𝐴 / 𝑥⦌𝐶 ↔ ∀𝑦(𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐵 → 𝑦 ∈ ⦋𝐴 / 𝑥⦌𝐶))
12	8, 10, 11	3bitr4g 317	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝐵 ⊆ 𝐶 ↔ ⦋𝐴 / 𝑥⦌𝐵 ⊆ ⦋𝐴 / 𝑥⦌𝐶))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   ∈ wcel 2146  [wsbc 3747  ⦋csb 3856   ⊆ wss 3908

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-fal 1583  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-nfc 2915  df-v 3460  df-sbc 3748  df-csb 3857  df-dif 3911  df-ss 3925  df-nul 4290

This theorem is used by:  sbcrel  5772  sbcfg  6710  csbfrecsg  8290  iuninc  32942  minregex  44301  brtrclfv2  44494  cotrclrcl  44509  sbcheg  44546