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Sounds like a good place to start.Should I use a K-map to see if the simplification exists.
oops, no you can't wire the outputs together to make an OR gate. I'll try the negative logic NAND approach...
LED layout:
0
1 2
3
4 5
6
Truth table:
ABCD 0123456
0 0000 1110111
1 0001 0010010
2 0010 1011101
3 0011 1011011
4 0100 0111010
5 0101 1101011
6 0110 1101111
7 0111 1010010
8 1000 1111111
9 1001 1111011
Equations:
0= !A.!B. C + !A.!B. !D + !A.B.C + !A.B. D + A
1= !A.!B.!C.!D + !A. B.!C + !A.B. !D + A
2= !A.!B + !A. B. C. D + !A.B.!C.!D + A
3= !A.!B. C + !A. B.!C + !A.B. !D + A
4= !A.!B. !D + !A. B. C.!D + A. !D
5= !A.!B.!C + !A.!B. D + !A.B + A
6= !A.!B. C + !A.!B.!D + !A.B.!C. D + !A.B.C.!D + A
Circuit:
_ _ _ _
A A B B C C D D
| | | | | | | |
| o | o o | | |-D\
| o | o | | | o-D \
| o o | o | | |-D OR-->0
| o o | | | o |-D /
o | | | | | | |-D/
| | | | | | | |
| o | o | o | o-D\
| o o | | o | |-D \OR-->1
| o o | | | o |-D /
o | | | | | | |-D/
| | | | | | | |
| o | o | | | |-D\
| o o | o | o |-D \OR-->2
| o o | o | o |-D /
o | | | | | | |-D/
| | | | | | | |
| o | o o | | |-D\
| o o | | o | |-D \OR-->3
| o o | | | | o-D /
o | | | | | | |-D/
| | | | | | | |
| o | o | | | o-D\
| o o | o | | o-D OR-->4
o | | | | | | o-D/
| | | | | | | |
| o | o | o | |-D\
| o | o | | o |-D \OR-->5
| o o | | | | |-D /
o | | | | | | |-D/
| | | | | | | |
| o | o o | | |-D\
| o | o | | | o-D \
| o o | | o o |-D OR-->6
| o o | o | | o-D /
o | | | | | | |-D/
| | | | | | | |
A--+--------
| NOR---> !(A+!B)
+- +-
NOR--+-------> !(A+B)
+- +-
| NOR---> !(!A+B)
B--+--------
0=( !B+ C+ D) . ( A+ B+ C+!D)
1=( !C+!D) . ( A+ B+ C+!D)
2=( !B+ C+!D) . ( !B+!C+ D)
3=( A+ B+ C ) . ( !B+!C+!D)
4=( !D) . ( !B+ C+ D)
5=( B+!C+ D)
6=( A+ B+ C+!D) . ( !B+ C+ D) . ( !B+!C+!D)