Theorem pltval3

Description: Alternate expression for the "less than" relation. ( dfpss3 analog.) (Contributed by NM, 4-Nov-2011)

	Ref	Expression
Hypotheses	pleval2.b	$$⊢B={\mathrm{Base}}_K$$
	pleval2.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
	pleval2.s	$$⊢\underset{\dot{}}{<}={<}_K$$
Assertion	pltval3	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(X\underset{\dot{}}{<}Y\leftrightarrow \left(X\underset{\dot{}}{\le }Y\wedge \neg Y\underset{\dot{}}{\le }X\right)\right)$$

Proof

Step	Hyp	Ref	Expression
1		pleval2.b	$$⊢B={\mathrm{Base}}_K$$
2		pleval2.l	$$⊢\underset{\dot{}}{\le }={\le }_K$$
3		pleval2.s	$$⊢\underset{\dot{}}{<}={<}_K$$
4	2 3	pltval	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(X\underset{\dot{}}{<}Y\leftrightarrow \left(X\underset{\dot{}}{\le }Y\wedge X\ne Y\right)\right)$$
5	1 2	posref	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\right)\to X\underset{\dot{}}{\le }X$$
6	5	3adant3	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to X\underset{\dot{}}{\le }X$$
7		breq1	$$⊢X=Y\to \left(X\underset{\dot{}}{\le }X\leftrightarrow Y\underset{\dot{}}{\le }X\right)$$
8	6 7	syl5ibcom	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(X=Y\to Y\underset{\dot{}}{\le }X\right)$$
9	8	adantr	$$⊢\left(\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\wedge X\underset{\dot{}}{\le }Y\right)\to \left(X=Y\to Y\underset{\dot{}}{\le }X\right)$$
10	1 2	posasymb	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(\left(X\underset{\dot{}}{\le }Y\wedge Y\underset{\dot{}}{\le }X\right)\leftrightarrow X=Y\right)$$
11	10	biimpd	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(\left(X\underset{\dot{}}{\le }Y\wedge Y\underset{\dot{}}{\le }X\right)\to X=Y\right)$$
12	11	expdimp	$$⊢\left(\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\wedge X\underset{\dot{}}{\le }Y\right)\to \left(Y\underset{\dot{}}{\le }X\to X=Y\right)$$
13	9 12	impbid	$$⊢\left(\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\wedge X\underset{\dot{}}{\le }Y\right)\to \left(X=Y\leftrightarrow Y\underset{\dot{}}{\le }X\right)$$
14	13	necon3abid	$$⊢\left(\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\wedge X\underset{\dot{}}{\le }Y\right)\to \left(X\ne Y\leftrightarrow \neg Y\underset{\dot{}}{\le }X\right)$$
15	14	pm5.32da	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(\left(X\underset{\dot{}}{\le }Y\wedge X\ne Y\right)\leftrightarrow \left(X\underset{\dot{}}{\le }Y\wedge \neg Y\underset{\dot{}}{\le }X\right)\right)$$
16	4 15	bitrd	$$⊢\left(K\in \mathrm{Poset}\wedge X\in B\wedge Y\in B\right)\to \left(X\underset{\dot{}}{<}Y\leftrightarrow \left(X\underset{\dot{}}{\le }Y\wedge \neg Y\underset{\dot{}}{\le }X\right)\right)$$