Metamath Proof Explorer


Theorem paddidm

Description: Projective subspace sum is idempotent. Part of Lemma 16.2 of MaedaMaeda p. 68. (Contributed by NM, 13-Jan-2012)

Ref Expression
Hypotheses paddidm.s S = PSubSp K
paddidm.p + ˙ = + 𝑃 K
Assertion paddidm K B X S X + ˙ X = X

Proof

Step Hyp Ref Expression
1 paddidm.s S = PSubSp K
2 paddidm.p + ˙ = + 𝑃 K
3 simpl K B X S K B
4 eqid Atoms K = Atoms K
5 4 1 psubssat K B X S X Atoms K
6 eqid K = K
7 eqid join K = join K
8 6 7 4 2 elpadd K B X Atoms K X Atoms K p X + ˙ X p X p X p Atoms K q X r X p K q join K r
9 3 5 5 8 syl3anc K B X S p X + ˙ X p X p X p Atoms K q X r X p K q join K r
10 pm1.2 p X p X p X
11 10 a1i K B X S p X p X p X
12 6 7 4 1 psubspi K B X S p Atoms K q X r X p K q join K r p X
13 12 3exp1 K B X S p Atoms K q X r X p K q join K r p X
14 13 imp4b K B X S p Atoms K q X r X p K q join K r p X
15 11 14 jaod K B X S p X p X p Atoms K q X r X p K q join K r p X
16 9 15 sylbid K B X S p X + ˙ X p X
17 16 ssrdv K B X S X + ˙ X X
18 4 2 sspadd1 K B X Atoms K X Atoms K X X + ˙ X
19 3 5 5 18 syl3anc K B X S X X + ˙ X
20 17 19 eqssd K B X S X + ˙ X = X