Theorem pjoml

Description: Subspace form of orthomodular law in the Hilbert lattice. Compare the orthomodular law in Theorem 2(ii) of Kalmbach p. 22. Derived using projections; compare omlsi . (Contributed by NM, 14-Jun-2006) (New usage is discouraged.)

	Ref	Expression
Assertion	pjoml	$$⊢\left(\left(A\in {\mathbf{C}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge \left(A\subseteq B\wedge B\cap \perp \left(A\right)=0_{\mathscr{H}}\right)\right)\to A=B$$

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		fveq2	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \perp \left(A\right)=\perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)$$
3	2	ineq2d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to B\cap \perp \left(A\right)=B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)$$
4	3	eqeq1d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(B\cap \perp \left(A\right)=0_{\mathscr{H}}\leftrightarrow B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)$$
5	1 4	anbi12d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(\left(A\subseteq B\wedge B\cap \perp \left(A\right)=0_{\mathscr{H}}\right)\leftrightarrow \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\wedge B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\right)$$
6		eqeq1	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(A=B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=B\right)$$
7	5 6	imbi12d	$$⊢A=if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\to \left(\left(\left(A\subseteq B\wedge B\cap \perp \left(A\right)=0_{\mathscr{H}}\right)\to A=B\right)\leftrightarrow \left(\left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\wedge B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=B\right)\right)$$
8		sseq2	$$⊢B=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)$$
9		ineq1	$$⊢B=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)$$
10	9	eqeq1d	$$⊢B=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\leftrightarrow if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)$$
11	8 10	anbi12d	$$⊢B=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(\left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\wedge B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\leftrightarrow \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\wedge if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\right)$$
12		eqeq2	$$⊢B=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=B\leftrightarrow if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)$$
13	11 12	imbi12d	$$⊢B=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\to \left(\left(\left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq B\wedge B\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=B\right)\leftrightarrow \left(\left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\wedge if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\right)\right)$$
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		h0elsh	$$⊢0_{\mathscr{H}}\in {\mathbf{S}}_{\mathscr{H}}$$
17	16	elimel	$$⊢if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\in {\mathbf{S}}_{\mathscr{H}}$$
18	15 17	pjomli	$$⊢\left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\subseteq if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\wedge if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)\cap \perp \left(if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)\right)=0_{\mathscr{H}}\right)\to if\left(A\in {\mathbf{C}}_{\mathscr{H}},A,0_{\mathscr{H}}\right)=if\left(B\in {\mathbf{S}}_{\mathscr{H}},B,0_{\mathscr{H}}\right)$$
19	7 13 18	dedth2h	$$⊢\left(A\in {\mathbf{C}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\to \left(\left(A\subseteq B\wedge B\cap \perp \left(A\right)=0_{\mathscr{H}}\right)\to A=B\right)$$
20	19	imp	$$⊢\left(\left(A\in {\mathbf{C}}_{\mathscr{H}}\wedge B\in {\mathbf{S}}_{\mathscr{H}}\right)\wedge \left(A\subseteq B\wedge B\cap \perp \left(A\right)=0_{\mathscr{H}}\right)\right)\to A=B$$