Theorem sn-subcl

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

	Ref	Expression
Assertion	sn-subcl	$$⊢\left(A\in ℂ\wedge B\in ℂ\right)\to A-B\in ℂ$$

Proof

Step	Hyp	Ref	Expression
1		subval	$$⊢\left(A\in ℂ\wedge B\in ℂ\right)\to A-B=\left(\iota x\in ℂ|B+x=A\right)$$
2		sn-subeu	$$⊢\left(B\in ℂ\wedge A\in ℂ\right)\to \exists !x\in ℂB+x=A$$
3	2	ancoms	$$⊢\left(A\in ℂ\wedge B\in ℂ\right)\to \exists !x\in ℂB+x=A$$
4		riotacl	$$⊢\exists !x\in ℂB+x=A\to \left(\iota x\in ℂ|B+x=A\right)\in ℂ$$
5	3 4	syl	$$⊢\left(A\in ℂ\wedge B\in ℂ\right)\to \left(\iota x\in ℂ|B+x=A\right)\in ℂ$$
6	1 5	eqeltrd	$$⊢\left(A\in ℂ\wedge B\in ℂ\right)\to A-B\in ℂ$$