Metamath Proof Explorer


Theorem paddasslem5

Description: Lemma for paddass . Show s =/= z by contradiction. (Contributed by NM, 8-Jan-2012)

Ref Expression
Hypotheses paddasslem.l ˙=K
paddasslem.j ˙=joinK
paddasslem.a A=AtomsK
Assertion paddasslem5 KHLrAxAyAzA¬r˙x˙yr˙y˙zs˙x˙ysz

Proof

Step Hyp Ref Expression
1 paddasslem.l ˙=K
2 paddasslem.j ˙=joinK
3 paddasslem.a A=AtomsK
4 breq1 s=zs˙x˙yz˙x˙y
5 4 biimpac s˙x˙ys=zz˙x˙y
6 eqid BaseK=BaseK
7 simpll1 KHLrAxAyAzAr˙y˙zz˙x˙yKHL
8 7 hllatd KHLrAxAyAzAr˙y˙zz˙x˙yKLat
9 simpll2 KHLrAxAyAzAr˙y˙zz˙x˙yrA
10 6 3 atbase rArBaseK
11 9 10 syl KHLrAxAyAzAr˙y˙zz˙x˙yrBaseK
12 simp32 KHLrAxAyAzAyA
13 12 ad2antrr KHLrAxAyAzAr˙y˙zz˙x˙yyA
14 6 3 atbase yAyBaseK
15 13 14 syl KHLrAxAyAzAr˙y˙zz˙x˙yyBaseK
16 simp33 KHLrAxAyAzAzA
17 16 ad2antrr KHLrAxAyAzAr˙y˙zz˙x˙yzA
18 6 3 atbase zAzBaseK
19 17 18 syl KHLrAxAyAzAr˙y˙zz˙x˙yzBaseK
20 6 2 latjcl KLatyBaseKzBaseKy˙zBaseK
21 8 15 19 20 syl3anc KHLrAxAyAzAr˙y˙zz˙x˙yy˙zBaseK
22 simp31 KHLrAxAyAzAxA
23 22 ad2antrr KHLrAxAyAzAr˙y˙zz˙x˙yxA
24 6 3 atbase xAxBaseK
25 23 24 syl KHLrAxAyAzAr˙y˙zz˙x˙yxBaseK
26 6 2 latjcl KLatxBaseKyBaseKx˙yBaseK
27 8 25 15 26 syl3anc KHLrAxAyAzAr˙y˙zz˙x˙yx˙yBaseK
28 simplr KHLrAxAyAzAr˙y˙zz˙x˙yr˙y˙z
29 1 2 3 hlatlej2 KHLxAyAy˙x˙y
30 7 23 13 29 syl3anc KHLrAxAyAzAr˙y˙zz˙x˙yy˙x˙y
31 simpr KHLrAxAyAzAr˙y˙zz˙x˙yz˙x˙y
32 6 1 2 latjle12 KLatyBaseKzBaseKx˙yBaseKy˙x˙yz˙x˙yy˙z˙x˙y
33 32 biimpd KLatyBaseKzBaseKx˙yBaseKy˙x˙yz˙x˙yy˙z˙x˙y
34 8 15 19 27 33 syl13anc KHLrAxAyAzAr˙y˙zz˙x˙yy˙x˙yz˙x˙yy˙z˙x˙y
35 30 31 34 mp2and KHLrAxAyAzAr˙y˙zz˙x˙yy˙z˙x˙y
36 6 1 8 11 21 27 28 35 lattrd KHLrAxAyAzAr˙y˙zz˙x˙yr˙x˙y
37 36 ex KHLrAxAyAzAr˙y˙zz˙x˙yr˙x˙y
38 5 37 syl5 KHLrAxAyAzAr˙y˙zs˙x˙ys=zr˙x˙y
39 38 expdimp KHLrAxAyAzAr˙y˙zs˙x˙ys=zr˙x˙y
40 39 necon3bd KHLrAxAyAzAr˙y˙zs˙x˙y¬r˙x˙ysz
41 40 exp31 KHLrAxAyAzAr˙y˙zs˙x˙y¬r˙x˙ysz
42 41 com23 KHLrAxAyAzAs˙x˙yr˙y˙z¬r˙x˙ysz
43 42 com24 KHLrAxAyAzA¬r˙x˙yr˙y˙zs˙x˙ysz
44 43 3imp2 KHLrAxAyAzA¬r˙x˙yr˙y˙zs˙x˙ysz