Deontic independence challenge

Which of these 128 combinations are satisfiable in Carmo and Jones style deontic logic? (1=true, 0=false) (Here 5(c) is the strong version.)
Some possibilities are ruled out by Observation 5.1.2 that says 5(bcde) imply 5(g); and Lemma II.2.2 that says 5(abcd) imply 5(f), formalized by RJ Reiff.

5(abcdefg) Model
0000000?
0000001?
0000010?
0000011?
0000100?
0000101?
0000110?
0000111?
0001000?
0001001?
0001010?
0001011?
0001100?
0001101?
0001110?
0001111?
0010000?
0010001?
0010010?
0010011?
0010100?
0010101?
0010110?
0010111?
0011000?
0011001?
0011010?
0011011?
0011100?
0011101?
0011110?
0011111?
0100000?
0100001?
0100010?
0100011?
0100100?
0100101?
0100110?
0100111?
0101000?
0101001?
0101010?
0101011?
0101100?
0101101?
0101110?
0101111?
0110000?
0110001?
0110010?
0110011?
0110100?
0110101?
0110110?
0110111?
0111000?
0111001?
0111010?
0111011?
0111100none; Obs. 5.1.2
0111101?
0111110none; Obs. 5.1.2
0111111see: A5_not_implied
1000000?
1000001?
1000010?
1000011?
1000100?
1000101?
1000110?
1000111?
1001000?
1001001?
1001010?
1001011?
1001100?
1001101?
1001110?
1001111?
1010000?
1010001?
1010010?
1010011?
1010100?
1010101?
1010110?
1010111?
1011000?
1011001?
1011010?
1011011?
1011100?
1011101?
1011110?
1011111see: B5_not_implied’
1100000?
1100001?
1100010?
1100011?
1100100?
1100101?
1100110?
1100111?
1101000?
1101001?
1101010?
1101011?
1101100see: C5_not_implied’
1101101?
1101110?
1101111?
1110000?
1110001?
1110010?
1110011?
1110100?
1110101see theorem: strong_do_not_imply_5d_or_5f
1110110?
1110111?
1111000none; see II.2.2
1111001none; see II.2.2
1111010?
1111011see theorem: do_not_imply_5e
1111100 impossible as 5(fg) follow from 5(abcde)
1111101 impossible as 5(fg) follow from 5(abcde)
1111110 impossible as 5(fg) follow from 5(abcde)
1111111 stayAlive, alive, noObligations

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