Theorem bj-gabss 36918

Description: Inclusion of generalized class abstractions. (Contributed by BJ, 4-Oct-2024.)

Assertion

Ref	Expression
bj-gabss	⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝐴 ∣ 𝑥 ∣ 𝜑} ⊆ {𝐵 ∣ 𝑥 ∣ 𝜓})

Proof of Theorem bj-gabss

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeq1 2739	. . . . . . . 8 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 ↔ 𝐵 = 𝑦))
2	1	biimpd 229	. . . . . . 7 ⊢ (𝐴 = 𝐵 → (𝐴 = 𝑦 → 𝐵 = 𝑦))
3	2	adantr 480	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝐴 = 𝑦 → 𝐵 = 𝑦))
4		simpr 484	. . . . . 6 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (𝜑 → 𝜓))
5	3, 4	anim12d 609	. . . . 5 ⊢ ((𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ((𝐴 = 𝑦 ∧ 𝜑) → (𝐵 = 𝑦 ∧ 𝜓)))
6	5	aleximi 1829	. . . 4 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → (∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
7	6	alrimiv 1925	. . 3 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
8		ss2ab 4072	. . 3 ⊢ ({𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)} ⊆ {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)} ↔ ∀𝑦(∃𝑥(𝐴 = 𝑦 ∧ 𝜑) → ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)))
9	7, 8	sylibr 234	. 2 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)} ⊆ {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)})
10		df-bj-gab 36917	. 2 ⊢ {𝐴 ∣ 𝑥 ∣ 𝜑} = {𝑦 ∣ ∃𝑥(𝐴 = 𝑦 ∧ 𝜑)}
11		df-bj-gab 36917	. 2 ⊢ {𝐵 ∣ 𝑥 ∣ 𝜓} = {𝑦 ∣ ∃𝑥(𝐵 = 𝑦 ∧ 𝜓)}
12	9, 10, 11	3sstr4g 4041	1 ⊢ (∀𝑥(𝐴 = 𝐵 ∧ (𝜑 → 𝜓)) → {𝐴 ∣ 𝑥 ∣ 𝜑} ⊆ {𝐵 ∣ 𝑥 ∣ 𝜓})

Syntax hints:   → wi 4   ∧ wa 395  ∀wal 1535   = wceq 1537  ∃wex 1776  {cab 2712   ⊆ wss 3963  {bj-cgab 36916

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1792  ax-4 1806  ax-5 1908  ax-6 1965  ax-7 2005  ax-8 2108  ax-9 2116  ax-10 2139  ax-11 2155  ax-12 2175  ax-ext 2706

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-ex 1777  df-nf 1781  df-sb 2063  df-clab 2713  df-cleq 2727  df-clel 2814  df-nfc 2890  df-ss 3980  df-bj-gab 36917

This theorem is referenced by:  bj-gabssd  36919