Metamath Proof Explorer


Theorem nosupbnd1lem2

Description: Lemma for nosupbnd1 . When there is no maximum, if any member of A is a prolongment of S , then so are all elements of A above it. (Contributed by Scott Fenton, 5-Dec-2021)

Ref Expression
Hypothesis nosupbnd1.1 S=ifxAyA¬x<syιxA|yA¬x<sydomιxA|yA¬x<sy2𝑜gy|uAydomuvA¬v<suusucy=vsucyιx|uAgdomuvA¬v<suusucg=vsucgug=x
Assertion nosupbnd1lem2 ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWdomS=S

Proof

Step Hyp Ref Expression
1 nosupbnd1.1 S=ifxAyA¬x<syιxA|yA¬x<sydomιxA|yA¬x<sy2𝑜gy|uAydomuvA¬v<suusucy=vsucyιx|uAgdomuvA¬v<suusucg=vsucgug=x
2 simp3rr ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sU¬W<sU
3 simp2l ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUANo
4 simp3rl ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWA
5 3 4 sseldd ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWNo
6 simp3ll ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUUA
7 3 6 sseldd ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUUNo
8 1 nosupno ANoAVSNo
9 8 3ad2ant2 ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUSNo
10 nodmon SNodomSOn
11 9 10 syl ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUdomSOn
12 sltres WNoUNodomSOnWdomS<sUdomSW<sU
13 5 7 11 12 syl3anc ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWdomS<sUdomSW<sU
14 2 13 mtod ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sU¬WdomS<sUdomS
15 simp3lr ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUUdomS=S
16 15 breq2d ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWdomS<sUdomSWdomS<sS
17 14 16 mtbid ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sU¬WdomS<sS
18 1 nosupbnd1lem1 ¬xAyA¬x<syANoAVWA¬S<sWdomS
19 4 18 syld3an3 ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sU¬S<sWdomS
20 noreson WNodomSOnWdomSNo
21 5 11 20 syl2anc ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWdomSNo
22 sltso <sOrNo
23 sotrieq2 <sOrNoWdomSNoSNoWdomS=S¬WdomS<sS¬S<sWdomS
24 22 23 mpan WdomSNoSNoWdomS=S¬WdomS<sS¬S<sWdomS
25 21 9 24 syl2anc ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWdomS=S¬WdomS<sS¬S<sWdomS
26 17 19 25 mpbir2and ¬xAyA¬x<syANoAVUAUdomS=SWA¬W<sUWdomS=S