Theorem invdisj

Description: If there is a function C ( y ) such that C ( y ) = x for all y e. B ( x ) , then the sets B ( x ) for distinct x e. A are disjoint. (Contributed by Mario Carneiro, 10-Dec-2016)

	Ref	Expression
Assertion	invdisj	$$⊢\forall x\in A\forall y\in BC=x\to \underset{x\in A}{Disj}B$$

Proof

Step	Hyp	Ref	Expression
1		nfra2w	$$⊢Ⅎy\forall x\in A\forall y\in BC=x$$
2		df-ral	$$⊢\forall x\in A\forall y\in BC=x\leftrightarrow \forall x\left(x\in A\to \forall y\in BC=x\right)$$
3		rsp	$$⊢\forall y\in BC=x\to \left(y\in B\to C=x\right)$$
4		eqcom	$$⊢C=x\leftrightarrow x=C$$
5	3 4	syl6ib	$$⊢\forall y\in BC=x\to \left(y\in B\to x=C\right)$$
6	5	imim2i	$$⊢\left(x\in A\to \forall y\in BC=x\right)\to \left(x\in A\to \left(y\in B\to x=C\right)\right)$$
7	6	impd	$$⊢\left(x\in A\to \forall y\in BC=x\right)\to \left(\left(x\in A\wedge y\in B\right)\to x=C\right)$$
8	7	alimi	$$⊢\forall x\left(x\in A\to \forall y\in BC=x\right)\to \forall x\left(\left(x\in A\wedge y\in B\right)\to x=C\right)$$
9	2 8	sylbi	$$⊢\forall x\in A\forall y\in BC=x\to \forall x\left(\left(x\in A\wedge y\in B\right)\to x=C\right)$$
10		mo2icl	$$⊢\forall x\left(\left(x\in A\wedge y\in B\right)\to x=C\right)\to {\exists }^{*}x\left(x\in A\wedge y\in B\right)$$
11	9 10	syl	$$⊢\forall x\in A\forall y\in BC=x\to {\exists }^{*}x\left(x\in A\wedge y\in B\right)$$
12	1 11	alrimi	$$⊢\forall x\in A\forall y\in BC=x\to \forall y{\exists }^{*}x\left(x\in A\wedge y\in B\right)$$
13		dfdisj2	$$⊢\underset{x\in A}{Disj}B\leftrightarrow \forall y{\exists }^{*}x\left(x\in A\wedge y\in B\right)$$
14	12 13	sylibr	$$⊢\forall x\in A\forall y\in BC=x\to \underset{x\in A}{Disj}B$$