Theorem shlej2

Description: Add disjunct to both sides of Hilbert subspace ordering. (Contributed by NM, 22-Jun-2004) (New usage is discouraged.)

	Ref	Expression
Assertion	shlej2	$$⊢\left(\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge A\subseteq B\right)\to C{\vee }_{\mathscr{H}}A\subseteq C{\vee }_{\mathscr{H}}B$$

Proof

Step	Hyp	Ref	Expression
1		shlej1	$$⊢\left(\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge A\subseteq B\right)\to A{\vee }_{\mathscr{H}}C\subseteq B{\vee }_{\mathscr{H}}C$$
2		shjcom	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\to A{\vee }_{\mathscr{H}}C=C{\vee }_{\mathscr{H}}A$$
3	2	3adant2	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\to A{\vee }_{\mathscr{H}}C=C{\vee }_{\mathscr{H}}A$$
4	3	adantr	$$⊢\left(\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge A\subseteq B\right)\to A{\vee }_{\mathscr{H}}C=C{\vee }_{\mathscr{H}}A$$
5		shjcom	$$⊢\left(B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\to B{\vee }_{\mathscr{H}}C=C{\vee }_{\mathscr{H}}B$$
6	5	3adant1	$$⊢\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\to B{\vee }_{\mathscr{H}}C=C{\vee }_{\mathscr{H}}B$$
7	6	adantr	$$⊢\left(\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge A\subseteq B\right)\to B{\vee }_{\mathscr{H}}C=C{\vee }_{\mathscr{H}}B$$
8	1 4 7	3sstr3d	$$⊢\left(\left(A\in {\mathbf{S}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\wedge C\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge A\subseteq B\right)\to C{\vee }_{\mathscr{H}}A\subseteq C{\vee }_{\mathscr{H}}B$$