Theorem supeq2 8914

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 3485	. . . 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 3407	. . . . . 6 ⊢ (𝐵 = 𝐶 → (∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧) ↔ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧)))
3	2	anbi2d 630	. . . . 5 ⊢ (𝐵 = 𝐶 → ((∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧)) ↔ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))))
4	3	rabbidv 3482	. . . 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 2858	. . 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 4854	. 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 8908	. 2 ⊢ sup(𝐴, 𝐵, 𝑅) = ∪ {𝑥 ∈ 𝐵 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐵 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))}
8		df-sup 8908	. 2 ⊢ sup(𝐴, 𝐶, 𝑅) = ∪ {𝑥 ∈ 𝐶 ∣ (∀𝑦 ∈ 𝐴 ¬ 𝑥𝑅𝑦 ∧ ∀𝑦 ∈ 𝐶 (𝑦𝑅𝑥 → ∃𝑧 ∈ 𝐴 𝑦𝑅𝑧))}
9	6, 7, 8	3eqtr4g 2883	1 ⊢ (𝐵 = 𝐶 → sup(𝐴, 𝐵, 𝑅) = sup(𝐴, 𝐶, 𝑅))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 398   = wceq 1537  ∀wral 3140  ∃wrex 3141  {crab 3144  ∪ cuni 4840   class class class wbr 5068  supcsup 8906

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-10 2145  ax-11 2161  ax-12 2177  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-or 844  df-tru 1540  df-ex 1781  df-nf 1785  df-sb 2070  df-clab 2802  df-cleq 2816  df-clel 2895  df-nfc 2965  df-ral 3145  df-rab 3149  df-v 3498  df-in 3945  df-ss 3954  df-uni 4841  df-sup 8908

This theorem is referenced by:  infeq2  8945