Theorem supeq3

Description: Equality theorem for supremum. (Contributed by Scott Fenton, 13-Jun-2018)

	Ref	Expression
Assertion	supeq3	⊢ ( 𝑅 = 𝑆 → sup ( 𝐴 , 𝐵 , 𝑅 ) = sup ( 𝐴 , 𝐵 , 𝑆 ) )

Proof

Step	Hyp	Ref	Expression
1		breq	⊢ ( 𝑅 = 𝑆 → ( 𝑥 𝑅 𝑦 ↔ 𝑥 𝑆 𝑦 ) )
2	1	notbid	⊢ ( 𝑅 = 𝑆 → ( ¬ 𝑥 𝑅 𝑦 ↔ ¬ 𝑥 𝑆 𝑦 ) )
3	2	ralbidv	⊢ ( 𝑅 = 𝑆 → ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑅 𝑦 ↔ ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑆 𝑦 ) )
4		breq	⊢ ( 𝑅 = 𝑆 → ( 𝑦 𝑅 𝑥 ↔ 𝑦 𝑆 𝑥 ) )
5		breq	⊢ ( 𝑅 = 𝑆 → ( 𝑦 𝑅 𝑧 ↔ 𝑦 𝑆 𝑧 ) )
6	5	rexbidv	⊢ ( 𝑅 = 𝑆 → ( ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ↔ ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) )
7	4 6	imbi12d	⊢ ( 𝑅 = 𝑆 → ( ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ) ↔ ( 𝑦 𝑆 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) ) )
8	7	ralbidv	⊢ ( 𝑅 = 𝑆 → ( ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ) ↔ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑆 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) ) )
9	3 8	anbi12d	⊢ ( 𝑅 = 𝑆 → ( ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ) ) ↔ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑆 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑆 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) ) ) )
10	9	rabbidv	⊢ ( 𝑅 = 𝑆 → { 𝑥 ∈ 𝐵 ∣ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ) ) } = { 𝑥 ∈ 𝐵 ∣ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑆 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑆 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) ) } )
11	10	unieqd	⊢ ( 𝑅 = 𝑆 → ∪ { 𝑥 ∈ 𝐵 ∣ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ) ) } = ∪ { 𝑥 ∈ 𝐵 ∣ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑆 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑆 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) ) } )
12		df-sup	⊢ sup ( 𝐴 , 𝐵 , 𝑅 ) = ∪ { 𝑥 ∈ 𝐵 ∣ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑅 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑅 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑅 𝑧 ) ) }
13		df-sup	⊢ sup ( 𝐴 , 𝐵 , 𝑆 ) = ∪ { 𝑥 ∈ 𝐵 ∣ ( ∀ 𝑦 ∈ 𝐴 ¬ 𝑥 𝑆 𝑦 ∧ ∀ 𝑦 ∈ 𝐵 ( 𝑦 𝑆 𝑥 → ∃ 𝑧 ∈ 𝐴 𝑦 𝑆 𝑧 ) ) }
14	11 12 13	3eqtr4g	⊢ ( 𝑅 = 𝑆 → sup ( 𝐴 , 𝐵 , 𝑅 ) = sup ( 𝐴 , 𝐵 , 𝑆 ) )