Home.
It is easy, but lengthy, to establish that a TQG over a small set is decomposable. What follows is a fully developed example of the procedure, which would also detect non-decomposability. It starts with the definition table of 4:14046:
| 4:14046 |
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| aaa = b | aab = a | aac = c | aad = d
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| baa = a | bab = b | bac = d | bad = c
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| caa = c | cab = d | cac = b | cad = a
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| daa = d | dab = c | dac = a | dad = b
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| aba = a | abb = b | abc = d | abd = c
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| bba = b | bbb = a | bbc = c | bbd = d
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| cba = d | cbb = c | cbc = a | cbd = b
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| dba = c | dbb = d | dbc = b | dbd = a
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| aca = d | acb = c | acc = a | acd = b
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| bca = c | bcb = d | bcc = b | bcd = a
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| cca = b | ccb = a | ccc = c | ccd = d
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| dca = a | dcb = b | dcc = d | dcd = c
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| ada = c | adb = d | adc = b | add = a
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| bda = d | bdb = c | bdc = a | bdd = b
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| cda = a | cdb = b | cdc = d | cdd = c
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| dda = b | ddb = a | ddc = c | ddd = d
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Recall that for sinisterior decomposability, there must exist F and G such that O (p, q, r) = F (G (p, q), r).
Hence expand the table into F and G notation:
| F (G (a, a), a) = b | F (G (a, a), b) = a
| F (G (a, a), c) = c | F (G (a, a), d) = d
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| F (G (b, a), a) = a | F (G (b, a), b) = b
| F (G (b, a), c) = d | F (G (b, a), d) = c
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| F (G (c, a), a) = c | F (G (c, a), b) = d
| F (G (c, a), c) = b | F (G (c, a), d) = a
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| F (G (d, a), a) = d | F (G (d, a), b) = c
| F (G (d, a), c) = a | F (G (d, a), d) = b
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| F (G (a, b), a) = a | F (G (a, b), b) = b
| F (G (a, b), c) = d | F (G (a, b), d) = c
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| F (G (b, b), a) = b | F (G (b, b), b) = a
| F (G (b, b), c) = c | F (G (b, b), d) = d
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| F (G (c, b), a) = d | F (G (c, b), b) = c
| F (G (c, b), c) = a | F (G (c, b), d) = b
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| F (G (d, b), a) = c | F (G (d, b), b) = d
| F (G (d, b), c) = b | F (G (d, b), d) = a
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| F (G (a, c), a) = d | F (G (a, c), b) = c
| F (G (a, c), c) = a | F (G (a, c), d) = b
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| F (G (b, c), a) = c | F (G (b, c), b) = d
| F (G (b, c), c) = b | F (G (b, c), d) = a
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| F (G (c, c), a) = b | F (G (c, c), b) = a
| F (G (c, c), c) = c | F (G (c, c), d) = d
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| F (G (d, c), a) = a | F (G (d, c), b) = b
| F (G (d, c), c) = d | F (G (d, c), d) = c
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| F (G (a, d), a) = c | F (G (a, d), b) = d
| F (G (a, d), c) = b | F (G (a, d), d) = a
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| F (G (b, d), a) = d | F (G (b, d), b) = c
| F (G (b, d), c) = a | F (G (b, d), d) = b
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| F (G (c, d), a) = a | F (G (c, d), b) = b
| F (G (c, d), c) = d | F (G (c, d), d) = c
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| F (G (d, d), a) = b | F (G (d, d), b) = a
| F (G (d, d), c) = c | F (G (d, d), d) = d
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To reduce congestion, write
- G (a, d) = w
- G (b, d) = x
- G (c, d) = y
- G (d, d) = z
| F (G (a, a), a) = b | F (G (a, a), b) = a
| F (G (a, a), c) = c | F (G (a, a), d) = d
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| F (G (b, a), a) = a | F (G (b, a), b) = b
| F (G (b, a), c) = d | F (G (b, a), d) = c
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| F (G (c, a), a) = c | F (G (c, a), b) = d
| F (G (c, a), c) = b | F (G (c, a), d) = a
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| F (G (d, a), a) = d | F (G (d, a), b) = c
| F (G (d, a), c) = a | F (G (d, a), d) = b
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| F (G (a, b), a) = a | F (G (a, b), b) = b
| F (G (a, b), c) = d | F (G (a, b), d) = c
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| F (G (b, b), a) = b | F (G (b, b), b) = a
| F (G (b, b), c) = c | F (G (b, b), d) = d
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| F (G (c, b), a) = d | F (G (c, b), b) = c
| F (G (c, b), c) = a | F (G (c, b), d) = b
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| F (G (d, b), a) = c | F (G (d, b), b) = d
| F (G (d, b), c) = b | F (G (d, b), d) = a
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| F (G (a, c), a) = d | F (G (a, c), b) = c
| F (G (a, c), c) = a | F (G (a, c), d) = b
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| F (G (b, c), a) = c | F (G (b, c), b) = d
| F (G (b, c), c) = b | F (G (b, c), d) = a
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| F (G (c, c), a) = b | F (G (c, c), b) = a
| F (G (c, c), c) = c | F (G (c, c), d) = d
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| F (G (d, c), a) = a | F (G (d, c), b) = b
| F (G (d, c), c) = d | F (G (d, c), d) = c
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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Now:
- From F (w, a) = c and F (G (b, c), a) = c, conclude G (b, c) = w.
- From F (x, a) = d and F (G (a, c), a) = d, conclude G (a, c) = x.
- From F (y, a) = a and F (G (d, c), a) = a, conclude G (d, c) = y.
- From F (z, a) = b and F (G (c, c), a) = b, conclude G (c, c) = z.
| F (G (a, a), a) = b | F (G (a, a), b) = a
| F (G (a, a), c) = c | F (G (a, a), d) = d
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| F (G (b, a), a) = a | F (G (b, a), b) = b
| F (G (b, a), c) = d | F (G (b, a), d) = c
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| F (G (c, a), a) = c | F (G (c, a), b) = d
| F (G (c, a), c) = b | F (G (c, a), d) = a
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| F (G (d, a), a) = d | F (G (d, a), b) = c
| F (G (d, a), c) = a | F (G (d, a), d) = b
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| F (G (a, b), a) = a | F (G (a, b), b) = b
| F (G (a, b), c) = d | F (G (a, b), d) = c
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| F (G (b, b), a) = b | F (G (b, b), b) = a
| F (G (b, b), c) = c | F (G (b, b), d) = d
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| F (G (c, b), a) = d | F (G (c, b), b) = c
| F (G (c, b), c) = a | F (G (c, b), d) = b
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| F (G (d, b), a) = c | F (G (d, b), b) = d
| F (G (d, b), c) = b | F (G (d, b), d) = a
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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Dismiss the last four rows as redundant. (Were they not redundant, they would contain a contradiction, and the decomposition would fail). Now:
- From F (w, a) = c and F (G (d, b), a) = c, conclude G (d, b) = w.
- From F (x, a) = d and F (G (c, b), a) = d, conclude G (c, b) = x.
- From F (y, a) = a and F (G (a, b), a) = a, conclude G (a, b) = y.
- From F (z, a) = b and F (G (b, b), a) = b, conclude G (b, b) = z.
| F (G (a, a), a) = b | F (G (a, a), b) = a
| F (G (a, a), c) = c | F (G (a, a), d) = d
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| F (G (b, a), a) = a | F (G (b, a), b) = b
| F (G (b, a), c) = d | F (G (b, a), d) = c
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| F (G (c, a), a) = c | F (G (c, a), b) = d
| F (G (c, a), c) = b | F (G (c, a), d) = a
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| F (G (d, a), a) = d | F (G (d, a), b) = c
| F (G (d, a), c) = a | F (G (d, a), d) = b
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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Again, dismiss the last four rows as redundant, and then:
- From F (w, a) = c and F (G (c, a), a) = c, conclude G (c, a) = w.
- From F (x, a) = d and F (G (d, a), a) = d, conclude G (d, a) = x.
- From F (y, a) = a and F (G (b, a), a) = a, conclude G (b, a) = y.
- From F (z, a) = b and F (G (a, a), a) = b, conclude G (a, a) = z.
| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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Once further, dismiss the last four rows as being redundant:
| F (z, a) = b | F (z, b) = a
| F (z, c) = c | F (z, d) = d
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| F (y, a) = a | F (y, b) = b
| F (y, c) = d | F (y, d) = c
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| F (w, a) = c | F (w, b) = d
| F (w, c) = b | F (w, d) = a
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| F (x, a) = d | F (x, b) = c
| F (x, c) = a | F (x, d) = b
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Gather the definitions of and conclusions about G:
| G (a, a) = z | G (a, b) = y
| G (a, c) = x | G (a, d) = w
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| G (b, a) = y | G (b, b) = z
| G (b, c) = w | G (b, d) = x
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| G (c, a) = w | G (c, b) = x
| G (c, c) = z | G (c, d) = y
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| G (d, a) = x | G (d, b) = w
| G (d, c) = y | G (d, d) = z
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With these tables that give F and G, the decomposition is established. Although w, x, y and z are distinct members of the set { a, b, c, d }, it does not matter which is which.