Metamath Proof Explorer


Theorem mdsymlem8

Description: Lemma for mdsymi . Lemma 4(ii) of Maeda p. 168. (Contributed by NM, 3-Jul-2004) (New usage is discouraged.)

Ref Expression
Hypotheses mdsymlem1.1 AC
mdsymlem1.2 BC
mdsymlem1.3 C=Ap
Assertion mdsymlem8 A0B0B𝑀*AA𝑀*B

Proof

Step Hyp Ref Expression
1 mdsymlem1.1 AC
2 mdsymlem1.2 BC
3 mdsymlem1.3 C=Ap
4 1 2 chjcomi AB=BA
5 4 sseq2i pABpBA
6 atelch qHAtomsqC
7 atelch rHAtomsrC
8 chjcom qCrCqr=rq
9 6 7 8 syl2an qHAtomsrHAtomsqr=rq
10 9 sseq2d qHAtomsrHAtomspqrprq
11 ancom qArBrBqA
12 11 a1i qHAtomsrHAtomsqArBrBqA
13 10 12 anbi12d qHAtomsrHAtomspqrqArBprqrBqA
14 13 2rexbiia qHAtomsrHAtomspqrqArBqHAtomsrHAtomsprqrBqA
15 rexcom qHAtomsrHAtomsprqrBqArHAtomsqHAtomsprqrBqA
16 14 15 bitri qHAtomsrHAtomspqrqArBrHAtomsqHAtomsprqrBqA
17 5 16 imbi12i pABqHAtomsrHAtomspqrqArBpBArHAtomsqHAtomsprqrBqA
18 17 ralbii pHAtomspABqHAtomsrHAtomspqrqArBpHAtomspBArHAtomsqHAtomsprqrBqA
19 18 a1i A0B0pHAtomspABqHAtomsrHAtomspqrqArBpHAtomspBArHAtomsqHAtomsprqrBqA
20 1 2 3 mdsymlem7 A0B0B𝑀*ApHAtomspABqHAtomsrHAtomspqrqArB
21 eqid Bp=Bp
22 2 1 21 mdsymlem7 B0A0A𝑀*BpHAtomspBArHAtomsqHAtomsprqrBqA
23 22 ancoms A0B0A𝑀*BpHAtomspBArHAtomsqHAtomsprqrBqA
24 19 20 23 3bitr4d A0B0B𝑀*AA𝑀*B