Theorem shjcom

Description: Commutative law for Hilbert lattice join of subspaces. (Contributed by NM, 22-Jun-2004) (New usage is discouraged.)

	Ref	Expression
Assertion	shjcom	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\to A{\vee }_{\mathscr{H}}B=B{\vee }_{\mathscr{H}}A$$

Proof

Step	Hyp	Ref	Expression
1		shjval	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\to A{\vee }_{\mathscr{H}}B=\perp \left(\perp \left(A\cup B\right)\right)$$
2		shjval	$$⊢\left(B\in {\mathbf{S}}_{\mathscr{H}}\wedge A\in {\mathbf{S}}_{\mathscr{H}}\right)\to B{\vee }_{\mathscr{H}}A=\perp \left(\perp \left(B\cup A\right)\right)$$
3	2	ancoms	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\to B{\vee }_{\mathscr{H}}A=\perp \left(\perp \left(B\cup A\right)\right)$$
4		uncom	$$⊢B\cup A=A\cup B$$
5	4	fveq2i	$$⊢\perp \left(B\cup A\right)=\perp \left(A\cup B\right)$$
6	5	fveq2i	$$⊢\perp \left(\perp \left(B\cup A\right)\right)=\perp \left(\perp \left(A\cup B\right)\right)$$
7	3 6	eqtrdi	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\to B{\vee }_{\mathscr{H}}A=\perp \left(\perp \left(A\cup B\right)\right)$$
8	1 7	eqtr4d	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\to A{\vee }_{\mathscr{H}}B=B{\vee }_{\mathscr{H}}A$$