✔ showing expression with variables substituted and result · helpdesk (published)

The result is that I much better understand what I'm asking for, and what the tools do. Maybe Cassette would do what I want. I wrote something pretty simple - which does the job ( abbreviated):

function run_identity(str)
    evp = eval ∘ Meta.parse
    evp("@show " * str)
end

# pvo=permutationsWithOrder(source-set, tuple-size)

# Associativity by exhaustion - feasible because 𝔹 is a small type.
pr("Associativity")
let
    comb = pwo(bools, 3)
    for (a, b, c) in comb
        run_identity("($a ⊻ $b) ⊻ $c == $a ⊻ ($b ⊻ $c)")
    end
end

Associativity
(one ⊻ zero) ⊻ zero == one ⊻ (zero ⊻ zero) = true
(zero ⊻ one) ⊻ one == zero ⊻ (one ⊻ one) = true
(one ⊻ zero) ⊻ one == one ⊻ (zero ⊻ one) = true
(zero ⊻ zero) ⊻ one == zero ⊻ (zero ⊻ one) = true
(one ⊻ one) ⊻ one == one ⊻ (one ⊻ one) = true
(zero ⊻ zero) ⊻ zero == zero ⊻ (zero ⊻ zero) = true
(zero ⊻ one) ⊻ zero == zero ⊻ (one ⊻ zero) = true
(one ⊻ one) ⊻ zero == one ⊻ (one ⊻ zero) = true

Commutativity
zero ⊻ one == one ⊻ zero = true
one ⊻ zero == zero ⊻ one = true
one ⊻ one == one ⊻ one = true
zero ⊻ zero == zero ⊻ zero = true

Self-inverting
zero ⊻ zero == 𝕫 = true
one ⊻ one == 𝕫 = true

Stream: helpdesk (published)

Topic: ✔ showing expression with variables substituted and result

Peter Goodall (Nov 27 2021 at 01:10):

Notification Bot (Nov 27 2021 at 01:14):