Theorem divsval

Description: The value of surreal division. (Contributed by Scott Fenton, 12-Mar-2025)

	Ref	Expression
Assertion	divsval	⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ( 𝐴 /su 𝐵 ) = ( ℩ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 𝐴 ) )

Proof

Step	Hyp	Ref	Expression
1		eldifsn	⊢ ( 𝐵 ∈ ( No ∖ { 0s } ) ↔ ( 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) )
2		eqeq2	⊢ ( 𝑦 = 𝐴 → ( ( 𝑧 ·s 𝑥 ) = 𝑦 ↔ ( 𝑧 ·s 𝑥 ) = 𝐴 ) )
3	2	riotabidv	⊢ ( 𝑦 = 𝐴 → ( ℩ 𝑥 ∈ No ( 𝑧 ·s 𝑥 ) = 𝑦 ) = ( ℩ 𝑥 ∈ No ( 𝑧 ·s 𝑥 ) = 𝐴 ) )
4		oveq1	⊢ ( 𝑧 = 𝐵 → ( 𝑧 ·s 𝑥 ) = ( 𝐵 ·s 𝑥 ) )
5	4	eqeq1d	⊢ ( 𝑧 = 𝐵 → ( ( 𝑧 ·s 𝑥 ) = 𝐴 ↔ ( 𝐵 ·s 𝑥 ) = 𝐴 ) )
6	5	riotabidv	⊢ ( 𝑧 = 𝐵 → ( ℩ 𝑥 ∈ No ( 𝑧 ·s 𝑥 ) = 𝐴 ) = ( ℩ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 𝐴 ) )
7		df-divs	⊢ /su = ( 𝑦 ∈ No , 𝑧 ∈ ( No ∖ { 0s } ) ↦ ( ℩ 𝑥 ∈ No ( 𝑧 ·s 𝑥 ) = 𝑦 ) )
8		riotaex	⊢ ( ℩ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 𝐴 ) ∈ V
9	3 6 7 8	ovmpo	⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ ( No ∖ { 0s } ) ) → ( 𝐴 /su 𝐵 ) = ( ℩ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 𝐴 ) )
10	1 9	sylan2br	⊢ ( ( 𝐴 ∈ No ∧ ( 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ) → ( 𝐴 /su 𝐵 ) = ( ℩ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 𝐴 ) )
11	10	3impb	⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ( 𝐴 /su 𝐵 ) = ( ℩ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 𝐴 ) )