Theorem cdleme31sn1c

Description: Part of proof of Lemma E in Crawley p. 113. (Contributed by NM, 1-Mar-2013)

	Ref	Expression
Hypotheses	cdleme31sn1c.g	$$⊢G=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(E\underset{\dot{}}{\vee }\left(\left(s\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
	cdleme31sn1c.i	$$⊢I=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=G\right)\right)$$
	cdleme31sn1c.n	$$⊢N=if\left(s\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right),I,D\right)$$
	cdleme31sn1c.y	$$⊢Y=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(E\underset{\dot{}}{\vee }\left(\left(R\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
	cdleme31sn1c.c	$$⊢C=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=Y\right)\right)$$
Assertion	cdleme31sn1c	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to ⦋R/s⦌N=C$$

Proof

Step	Hyp	Ref	Expression
1		cdleme31sn1c.g	$$⊢G=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(E\underset{\dot{}}{\vee }\left(\left(s\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
2		cdleme31sn1c.i	$$⊢I=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=G\right)\right)$$
3		cdleme31sn1c.n	$$⊢N=if\left(s\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right),I,D\right)$$
4		cdleme31sn1c.y	$$⊢Y=\left(P\underset{\dot{}}{\vee }Q\right)\underset{\dot{}}{\wedge }\left(E\underset{\dot{}}{\vee }\left(\left(R\underset{\dot{}}{\vee }t\right)\underset{\dot{}}{\wedge }W\right)\right)$$
5		cdleme31sn1c.c	$$⊢C=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=Y\right)\right)$$
6		eqid	$$⊢\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\right)=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\right)$$
7	2 3 6	cdleme31sn1	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to ⦋R/s⦌N=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\right)$$
8	1 4	cdleme31se	$$⊢R\in A\to ⦋R/s⦌G=Y$$
9	8	adantr	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to ⦋R/s⦌G=Y$$
10	9	eqeq2d	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \left(y=⦋R/s⦌G\leftrightarrow y=Y\right)$$
11	10	imbi2d	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \left(\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\leftrightarrow \left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=Y\right)\right)$$
12	11	ralbidv	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \left(\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\leftrightarrow \forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=Y\right)\right)$$
13	12	riotabidv	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\right)=\left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=Y\right)\right)$$
14	13 5	eqtr4di	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to \left(\iota y\in B|\forall t\in A\left(\left(\neg t\underset{\dot{}}{\le }W\wedge \neg t\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to y=⦋R/s⦌G\right)\right)=C$$
15	7 14	eqtrd	$$⊢\left(R\in A\wedge R\underset{\dot{}}{\le }\left(P\underset{\dot{}}{\vee }Q\right)\right)\to ⦋R/s⦌N=C$$