Theorem divsclw

Description: Weak division closure law. (Contributed by Scott Fenton, 12-Mar-2025)

	Ref	Expression
Assertion	divsclw	⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ( 𝐴 /su 𝐵 ) ∈ No )

Proof

Step	Hyp	Ref	Expression
1		divsval	⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) → ( 𝐴 /su 𝐵 ) = ( ℩ 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 ) )
2	1	adantr	⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ( 𝐴 /su 𝐵 ) = ( ℩ 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 ) )
3		3anrot	⊢ ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ↔ ( 𝐵 ∈ No ∧ 𝐵 ≠ 0s ∧ 𝐴 ∈ No ) )
4		noreceuw	⊢ ( ( ( 𝐵 ∈ No ∧ 𝐵 ≠ 0s ∧ 𝐴 ∈ No ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ∃! 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 )
5	3 4	sylanb	⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ∃! 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 )
6		riotacl	⊢ ( ∃! 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 → ( ℩ 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 ) ∈ No )
7	5 6	syl	⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ( ℩ 𝑦 ∈ No ( 𝐵 ·s 𝑦 ) = 𝐴 ) ∈ No )
8	2 7	eqeltrd	⊢ ( ( ( 𝐴 ∈ No ∧ 𝐵 ∈ No ∧ 𝐵 ≠ 0s ) ∧ ∃ 𝑥 ∈ No ( 𝐵 ·s 𝑥 ) = 1s ) → ( 𝐴 /su 𝐵 ) ∈ No )