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§1C1 Four inputs. What follows is an example of why four-input operations are not studied in this report: they are unwieldy.

For comparison, here are the 5 = 3! − 1 ways that commutativity could be defined for a three-input operation O:

 one input is fixed: O (p, q, r) = O (q, p, r) O (p, q, r) = O (r, q, p) O (p, q, r) = O (p, r, q) no input is fixed: O (p, q, r) = O (r, p, q) O (p, q, r) = O (q, r, p)

Here are the 23 = 4! − 1 ways that commutativity could be defined for a four-input operation N:

 two inputs are fixed: N (p, q, r, s) = N (p, q, s, r) N (p, q, r, s) = N (p, r, q, s) N (p, q, r, s) = N (p, s, r, q) N (p, q, r, s) = N (q, p, r, s) N (p, q, r, s) = N (r, q, p, s) N (p, q, r, s) = N (s, q, r, p) one input is fixed: N (p, q, r, s) = N (p, r, s, q) N (p, q, r, s) = N (p, s, q, r) N (p, q, r, s) = N (q, r, p, s) N (p, q, r, s) = N (q, s, r, p) N (p, q, r, s) = N (r, p, q, s) N (p, q, r, s) = N (r, q, s, p) N (p, q, r, s) = N (s, p, r, q) N (p, q, r, s) = N (s, q, p, r) no input is fixed: N (p, q, r, s) = N (q, p, s, r) N (p, q, r, s) = N (q, r, s, p) N (p, q, r, s) = N (q, s, p, r) N (p, q, r, s) = N (r, p, s, q) N (p, q, r, s) = N (r, s, p, q) N (p, q, r, s) = N (r, s, q, p) N (p, q, r, s) = N (s, p, q, r) N (p, q, r, s) = N (s, r, p, q) N (p, q, r, s) = N (s, r, q, p)

23 is too many formulas for most practical purposes.

§1C2 Majority operation. Although not a TQG, the simplest and most familiar nondecomposable ternary function is the majority operation M over values a and b, defined by:

1. M (a, a, a) = a
2. M (a, a, b) = a
3. M (a, b, a) = a
4. M (a, b, b) = b
5. M (b, a, a) = a
6. M (b, a, b) = b
7. M (b, b, a) = b
8. M (b, b, b) = b

Suppose that there were a decomposition M (p, q, r) = F (p, G (q, r)). Then we would have:

1. F (a, G (a, a)) = a
2. F (a, G (a, b)) = a
3. F (a, G (b, a)) = a
4. F (a, G (b, b)) = b
5. F (b, G (a, a)) = a
6. F (b, G (a, b)) = b
7. F (b, G (b, a)) = b
8. F (b, G (b, b)) = b

Observe:

• From 5 and 6, G (a, a) ≠ G (a, b).
• From 2 and 4, G (a, b) ≠ G (b, b).
• From 1 and 4, G (b, b) ≠ G (a, a).
Thus G (a, a), G (a, b) and G (b, b) must be three distinct values. However there are only two values to choose from, namely a and b, so decomposition fails.

§1C3 Cross product. There is a recognized ternary operation that delivers a result of the same data type as its inputs; it is a particular four-dimensional generalization of the cross product.

Let A be the vector [ a0, a1, a2, a3 ]; similarly for B and C. Also let W = { W0, W1, W2, W3 } be an orthonormal basis of an 4-dimensional vector space. Then it becomes convenient to define this cross product as the determinant of a formal matrix. (This report uses an HTML table to display the matrix.)

A × B × C = det
 a0 a1 a2 a3 b0 b1 b2 b3 c0 c1 c2 c3 W0 W1 W2 W3

To write this another way, expand the determinant. Let R = A × B × C. Then:

 r0 = + a3b2c1 − a2b3c1 + a2b1c3 − a3b1c2 + a1b3c2 − a1b2c3 r1 = + a2b3c0 − a3b2c0 + a3b0c2 − a2b0c3 + a0b2c3 − a0b3c2 r2 = + a1b0c3 − a0b1c3 + a0b3c1 − a1b3c0 + a3b1c0 − a3b0c1 r3 = + a0b1c2 − a1b0c2 + a1b2c0 − a0b2c1 + a2b0c1 − a2b1c0