Metamath Proof Explorer


Theorem 3oalem5

Description: Lemma for 3OA (weak) orthoarguesian law. (Contributed by NM, 19-Oct-1999) (New usage is discouraged.)

Ref Expression
Hypotheses 3oa.1 A C
3oa.2 B C
3oa.3 C C
3oa.4 R = B B A
3oa.5 S = C C A
Assertion 3oalem5 B + R C + S = B R C S

Proof

Step Hyp Ref Expression
1 3oa.1 A C
2 3oa.2 B C
3 3oa.3 C C
4 3oa.4 R = B B A
5 3oa.5 S = C C A
6 4 3oalem4 R B
7 2 choccli B C
8 2 1 chjcli B A C
9 7 8 chincli B B A C
10 4 9 eqeltri R C
11 10 2 osumi R B R + B = R B
12 6 11 ax-mp R + B = R B
13 2 chshii B S
14 10 chshii R S
15 13 14 shscomi B + R = R + B
16 2 10 chjcomi B R = R B
17 12 15 16 3eqtr4i B + R = B R
18 5 3oalem4 S C
19 3 choccli C C
20 3 1 chjcli C A C
21 19 20 chincli C C A C
22 5 21 eqeltri S C
23 22 3 osumi S C S + C = S C
24 18 23 ax-mp S + C = S C
25 3 chshii C S
26 22 chshii S S
27 25 26 shscomi C + S = S + C
28 3 22 chjcomi C S = S C
29 24 27 28 3eqtr4i C + S = C S
30 17 29 ineq12i B + R C + S = B R C S