[CG] Re: Functional Logic • Inquiry and Analogy

Cf: Functional Logic • Inquiry and Analogy • 15 https://inquiryintoinquiry.com/2022/05/20/functional-logic-inquiry-and-anal… Inquiry and Analogy • Measure for Measure https://oeis.org/wiki/Functional_Logic_%E2%80%A2_Inquiry_and_Analogy#Measur… All, Let us define two families of measures, α_i, β_i : (B × B → B) → B for i = 0 to 15, by means of the following equations: • α_i f = Υ(f_i, f) = Υ(f_i ⇒ f), • β_i f = Υ(f, f_i) = Υ(f ⇒ f_i). Table 14 shows the value of each α_i on each of the 16 boolean functions f : B × B → B. In terms of the implication ordering on the 16 functions, α_i f = 1 says that f is “above or identical to” f_i in the implication lattice, that is, f ≥ f_i in the implication ordering. Table 14. Qualifiers of the Implication Ordering α_i f = Υ(f_i, f) https://inquiryintoinquiry.files.wordpress.com/2022/05/qualifiers-of-implic… Table 15 shows the value of each β_i on each of the 16 boolean functions f : B × B → B. In terms of the implication ordering on the 16 functions, β_i f = 1 says that f is “below or identical to” f_i in the implication lattice, that is, f ≤ f_i in the implication ordering. Table 15. Qualifiers of the Implication Ordering β_i f = Υ(f, f_i) https://inquiryintoinquiry.files.wordpress.com/2022/05/qualifiers-of-implic… Applied to a given proposition f, the qualifiers α_i and β_i tell whether f is above f_i or below f_i, respectively, in the implication ordering. By way of example, let us trace the effects of several such measures, namely, those which occupy the limiting positions in the Tables. • α₀f = 1 iff f₀ ⇒ f iff 0 ⇒ f, hence α₀f = 1 for all f. • α₁₅f = 1 iff f₁₅ ⇒ f iff 1 ⇒ f, hence α₁₅f = 1 iff f = 1. • β₀f = 1 iff f ⇒ f₀ iff f ⇒ 0, hence β₀f = 1 iff f = 0. • β₁₅f = 1 iff f ⇒ f₁₅ iff f ⇒ 1, hence β₁₅f = 1 for all f. Expressed in terms of the propositional forms they value positively, α₀ = β₁₅ is a totally indiscriminate measure, accepting all propositions f : B × B → B, whereas α₁₅ and β₀ are measures which value the constant propositions 1 : B × B → B and 0 : B × B → B, respectively, above all others. Regards, Jon