Theorem sn-subcl

Description: subcl without ax-mulcom . (Contributed by SN, 5-May-2024)

	Ref	Expression
Assertion	sn-subcl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )

Proof

Step	Hyp	Ref	Expression
1		subval	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) = ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) )
2		sn-subeu	⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 )
3	2	ancoms	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 )
4		riotacl	⊢ ( ∃! 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 → ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) ∈ ℂ )
5	3 4	syl	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ℩ 𝑥 ∈ ℂ ( 𝐵 + 𝑥 ) = 𝐴 ) ∈ ℂ )
6	1 5	eqeltrd	⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )