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Description: Swap 2nd and 3rd members of lattice join. Lemma 2.2 in MegPav2002 p. 362. (Contributed by NM, 23-Jun-2012)
| Ref | Expression | ||
|---|---|---|---|
| Hypotheses | latjass.b | |- B = ( Base ` K ) |
|
| latjass.j | |- .\/ = ( join ` K ) |
||
| Assertion | latj31 | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( ( X .\/ Y ) .\/ Z ) = ( ( Z .\/ Y ) .\/ X ) ) |
| Step | Hyp | Ref | Expression |
|---|---|---|---|
| 1 | latjass.b | |- B = ( Base ` K ) |
|
| 2 | latjass.j | |- .\/ = ( join ` K ) |
|
| 3 | simpl | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> K e. Lat ) |
|
| 4 | simpr3 | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> Z e. B ) |
|
| 5 | simpr1 | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> X e. B ) |
|
| 6 | simpr2 | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> Y e. B ) |
|
| 7 | 1 2 | latj12 | |- ( ( K e. Lat /\ ( Z e. B /\ X e. B /\ Y e. B ) ) -> ( Z .\/ ( X .\/ Y ) ) = ( X .\/ ( Z .\/ Y ) ) ) |
| 8 | 3 4 5 6 7 | syl13anc | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( Z .\/ ( X .\/ Y ) ) = ( X .\/ ( Z .\/ Y ) ) ) |
| 9 | 1 2 | latjcl | |- ( ( K e. Lat /\ X e. B /\ Y e. B ) -> ( X .\/ Y ) e. B ) |
| 10 | 9 | 3adant3r3 | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( X .\/ Y ) e. B ) |
| 11 | 1 2 | latjcom | |- ( ( K e. Lat /\ ( X .\/ Y ) e. B /\ Z e. B ) -> ( ( X .\/ Y ) .\/ Z ) = ( Z .\/ ( X .\/ Y ) ) ) |
| 12 | 3 10 4 11 | syl3anc | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( ( X .\/ Y ) .\/ Z ) = ( Z .\/ ( X .\/ Y ) ) ) |
| 13 | 1 2 | latjcl | |- ( ( K e. Lat /\ Z e. B /\ Y e. B ) -> ( Z .\/ Y ) e. B ) |
| 14 | 3 4 6 13 | syl3anc | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( Z .\/ Y ) e. B ) |
| 15 | 1 2 | latjcom | |- ( ( K e. Lat /\ ( Z .\/ Y ) e. B /\ X e. B ) -> ( ( Z .\/ Y ) .\/ X ) = ( X .\/ ( Z .\/ Y ) ) ) |
| 16 | 3 14 5 15 | syl3anc | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( ( Z .\/ Y ) .\/ X ) = ( X .\/ ( Z .\/ Y ) ) ) |
| 17 | 8 12 16 | 3eqtr4d | |- ( ( K e. Lat /\ ( X e. B /\ Y e. B /\ Z e. B ) ) -> ( ( X .\/ Y ) .\/ Z ) = ( ( Z .\/ Y ) .\/ X ) ) |