Theorem eqfvelsetpreimafv

Description: If an element of the domain of the function has the same function value as an element of the preimage of a function value, then it is an element of the same preimage. (Contributed by AV, 9-Mar-2024)

	Ref	Expression
Hypothesis	setpreimafvex.p	$$⊢P=\left\{z|\exists x\in Az=F^{-1}\left[\left\{F\left(x\right)\right\}\right]\right\}$$
Assertion	eqfvelsetpreimafv	$$⊢\left(FFnA\wedge S\in P\wedge X\in S\right)\to \left(\left(Y\in A\wedge F\left(Y\right)=F\left(X\right)\right)\to Y\in S\right)$$

Proof

Step	Hyp	Ref	Expression
1		setpreimafvex.p	$$⊢P=\left\{z|\exists x\in Az=F^{-1}\left[\left\{F\left(x\right)\right\}\right]\right\}$$
2	1	elsetpreimafvbi	$$⊢\left(FFnA\wedge S\in P\wedge X\in S\right)\to \left(Y\in S\leftrightarrow \left(Y\in A\wedge F\left(Y\right)=F\left(X\right)\right)\right)$$
3	2	biimprd	$$⊢\left(FFnA\wedge S\in P\wedge X\in S\right)\to \left(\left(Y\in A\wedge F\left(Y\right)=F\left(X\right)\right)\to Y\in S\right)$$