Theorem pjoml2

Description: Variation of orthomodular law. Definition in Kalmbach p. 22. (Contributed by NM, 13-Jun-2006) (New usage is discouraged.)

	Ref	Expression
Assertion	pjoml2	$$⊢\left(A\in {\mathbf{C}}_{\mathscr{H}}\wedge B\in {\mathbf{C}}_{\mathscr{H}}\wedge A\subseteq B\right)\to A{\vee }_{\mathscr{H}}\left(\perp \left(A\right)\cap B\right)=B$$

Proof

Step	Hyp	Ref	Expression
1		sseq1	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(A\subseteq B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\right)$$
2		id	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)$$
3		fveq2	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \perp \left(A\right)=\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)$$
4	3	ineq1d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \perp \left(A\right)\cap B=\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B$$
5	2 4	oveq12d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to A{\vee }_{\mathscr{H}}\left(\perp \left(A\right)\cap B\right)=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B\right)$$
6	5	eqeq1d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(A{\vee }_{\mathscr{H}}\left(\perp \left(A\right)\cap B\right)=B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B\right)=B\right)$$
7	1 6	imbi12d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(\left(A\subseteq B\to A{\vee }_{\mathscr{H}}\left(\perp \left(A\right)\cap B\right)=B\right)\leftrightarrow \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B\right)=B\right)\right)$$
8		sseq2	$$⊢B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)$$
9		ineq2	$$⊢B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B=\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)$$
10	9	oveq2d	$$⊢B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B\right)=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)$$
11		id	$$⊢B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)$$
12	10 11	eqeq12d	$$⊢B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B\right)=B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)$$
13	8 12	imbi12d	$$⊢B=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(\left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap B\right)=B\right)\leftrightarrow \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)\right)$$
14		h0elch	$$⊢0_{\mathscr{H}}\in {\mathbf{C}}_{\mathscr{H}}$$
15	14	elimel	$$⊢if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\in {\mathbf{C}}_{\mathscr{H}}$$
16	14	elimel	$$⊢if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\in {\mathbf{C}}_{\mathscr{H}}$$
17	15 16	pjoml2i	$$⊢if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right){\vee }_{\mathscr{H}}\left(\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)\cap if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)=if\left(B\in {\mathbf{C}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)$$
18	7 13 17	dedth2h	$$⊢\left(A\in {\mathbf{C}}_{\mathscr{H}}\wedge B\in {\mathbf{C}}_{\mathscr{H}}\right)\to \left(A\subseteq B\to A{\vee }_{\mathscr{H}}\left(\perp \left(A\right)\cap B\right)=B\right)$$
19	18	3impia	$$⊢\left(A\in {\mathbf{C}}_{\mathscr{H}}\wedge B\in {\mathbf{C}}_{\mathscr{H}}\wedge A\subseteq B\right)\to A{\vee }_{\mathscr{H}}\left(\perp \left(A\right)\cap B\right)=B$$