Theorem supclt

Description: Closure of supremum. (Contributed by Jeff Madsen, 2-Sep-2009)

	Ref	Expression
Assertion	supclt	$$⊢\left(R\mathrm{Or}A\wedge \exists x\in A\left(\forall y\in B\neg xRy\wedge \forall y\in A\left(yRx\to \exists z\in ByRz\right)\right)\right)\to sup\left(B,A,R\right)\in A$$

Proof

Step	Hyp	Ref	Expression
1		simpl	$$⊢\left(R\mathrm{Or}A\wedge \exists x\in A\left(\forall y\in B\neg xRy\wedge \forall y\in A\left(yRx\to \exists z\in ByRz\right)\right)\right)\to R\mathrm{Or}A$$
2		simpr	$$⊢\left(R\mathrm{Or}A\wedge \exists x\in A\left(\forall y\in B\neg xRy\wedge \forall y\in A\left(yRx\to \exists z\in ByRz\right)\right)\right)\to \exists x\in A\left(\forall y\in B\neg xRy\wedge \forall y\in A\left(yRx\to \exists z\in ByRz\right)\right)$$
3	1 2	supcl	$$⊢\left(R\mathrm{Or}A\wedge \exists x\in A\left(\forall y\in B\neg xRy\wedge \forall y\in A\left(yRx\to \exists z\in ByRz\right)\right)\right)\to sup\left(B,A,R\right)\in A$$