Theorem supeq2 9407

Description: Equality theorem for supremum. (Contributed by Jeff Madsen, 2-Sep-2009.)

Assertion

Ref	Expression
supeq2	⊢ (𝐵 = 𝐶 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐶, 𝑅))

Proof of Theorem supeq2

Dummy variables 𝑥 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		rabeq 3437	. . . 4 ⊢ (𝐵 = 𝐶 → {𝑥 ∈ 𝐵 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))} = {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))})
2		raleq 3326	. . . . . 6 ⊢ (𝐵 = 𝐶 → (∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧) ↔ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧)))
3	2	anbi2d 641	. . . . 5 ⊢ (𝐵 = 𝐶 → ((∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧)) ↔ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))))
4	3	rabbidv 3430	. . . 4 ⊢ (𝐵 = 𝐶 → {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))} = {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))})
5	1, 4	eqtrd 2804	. . 3 ⊢ (𝐵 = 𝐶 → {𝑥 ∈ 𝐵 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))} = {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))})
6	5	unieqd 4889	. 2 ⊢ (𝐵 = 𝐶 → ∪ {𝑥 ∈ 𝐵 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))} = ∪ {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))})
7		df-sup 9401	. 2 ⊢ sup(𝐴, 𝐵, 𝑅) = ∪ {𝑥 ∈ 𝐵 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))}
8		df-sup 9401	. 2 ⊢ sup(𝐴, 𝐶, 𝑅) = ∪ {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))}
9	6, 7, 8	3eqtr4g 2829	1 ⊢ (𝐵 = 𝐶 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐶, 𝑅))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 400   = wceq 1567  ∀wral 3085  ∃wrex 3095  {crab 3423  ∪ cuni 4876   class class class wbr 5113  supcsup 9399

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-8 2151  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-clel 2844  df-ral 3086  df-rex 3096  df-rab 3424  df-v 3465  df-ss 3930  df-uni 4877  df-sup 9401

This theorem is referenced by:  infeq2  9439