Metamath Proof Explorer

Theorem atmod1i1

Description: Version of modular law pmod1i that holds in a Hilbert lattice, when one element is an atom. (Contributed by NM, 11-May-2012) (Revised by Mario Carneiro, 10-May-2013)

Ref Expression
Hypotheses atmod.b 
|- B = ( Base ` K )
atmod.l 
|- .<_ = ( le ` K )
atmod.j 
|- .\/ = ( join ` K )
atmod.m 
|- ./\ = ( meet ` K )
atmod.a 
|- A = ( Atoms ` K )
Assertion atmod1i1 
|- ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) /\ P .<_ Y ) -> ( P .\/ ( X ./\ Y ) ) = ( ( P .\/ X ) ./\ Y ) )

Proof

Step Hyp Ref Expression
1 atmod.b 
 |-  B = ( Base ` K )
2 atmod.l 
 |-  .<_ = ( le ` K )
3 atmod.j 
 |-  .\/ = ( join ` K )
4 atmod.m 
 |-  ./\ = ( meet ` K )
5 atmod.a 
 |-  A = ( Atoms ` K )
6 simpl 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) ) -> K e. HL )
7 simpr2 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) ) -> X e. B )
8 simpr1 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) ) -> P e. A )
9 eqid 
 |-  ( pmap ` K ) = ( pmap ` K )
10 eqid 
 |-  ( +P ` K ) = ( +P ` K )
11 1 3 5 9 10 pmapjat2 
 |-  ( ( K e. HL /\ X e. B /\ P e. A ) -> ( ( pmap ` K ) ` ( P .\/ X ) ) = ( ( ( pmap ` K ) ` P ) ( +P ` K ) ( ( pmap ` K ) ` X ) ) )
12 6 7 8 11 syl3anc 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) ) -> ( ( pmap ` K ) ` ( P .\/ X ) ) = ( ( ( pmap ` K ) ` P ) ( +P ` K ) ( ( pmap ` K ) ` X ) ) )
13 1 5 atbase 
 |-  ( P e. A -> P e. B )
14 1 2 3 4 9 10 hlmod1i 
 |-  ( ( K e. HL /\ ( P e. B /\ X e. B /\ Y e. B ) ) -> ( ( P .<_ Y /\ ( ( pmap ` K ) ` ( P .\/ X ) ) = ( ( ( pmap ` K ) ` P ) ( +P ` K ) ( ( pmap ` K ) ` X ) ) ) -> ( ( P .\/ X ) ./\ Y ) = ( P .\/ ( X ./\ Y ) ) ) )
15 13 14 syl3anr1 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) ) -> ( ( P .<_ Y /\ ( ( pmap ` K ) ` ( P .\/ X ) ) = ( ( ( pmap ` K ) ` P ) ( +P ` K ) ( ( pmap ` K ) ` X ) ) ) -> ( ( P .\/ X ) ./\ Y ) = ( P .\/ ( X ./\ Y ) ) ) )
16 12 15 mpan2d 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) ) -> ( P .<_ Y -> ( ( P .\/ X ) ./\ Y ) = ( P .\/ ( X ./\ Y ) ) ) )
17 16 3impia 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) /\ P .<_ Y ) -> ( ( P .\/ X ) ./\ Y ) = ( P .\/ ( X ./\ Y ) ) )
18 17 eqcomd 
 |-  ( ( K e. HL /\ ( P e. A /\ X e. B /\ Y e. B ) /\ P .<_ Y ) -> ( P .\/ ( X ./\ Y ) ) = ( ( P .\/ X ) ./\ Y ) )