Three-Argument Operations and Four-Argument Operations 1
Michał Muzalewski, Wojciech Skaba · 1991
We now state the proposition (1) For every function f and for arbitrary a, b, c holds f(a, b, c) = f(〈a, b, c〉). For simplicity we adopt the following rules: A, B, C, D are non-empty sets, a is an element of A, b is an element of B, and c is an element of C. Let us consider A, B, C, D, and let f be a function from [:A, B, C :] into D, and let us consider a, b, c. Then f(a, b, c) is an element of D. We adopt the following rules: X, Y , Z denote sets, T denotes a non-empty set, and x, y, z are arbitrary. One can prove the following propositions: (2) For all functions f1, f2 from [:X, Y, Z :] into T such that T 6= ∅ and for all x, y, z such that x ∈ X and y ∈ Y and z ∈ Z holds f1(〈x, y, z〉) = f2(〈x, y, z〉) holds f1 = f2. (3) For all functions f1, f2 from [:A, B, C :] into D such that for all a, b, c holds f1(〈a, b, c〉) = f2(〈a, b, c〉) holds f1 = f2.