218 lines
No EOL
6 KiB
Text
218 lines
No EOL
6 KiB
Text
#set document(title: "Homework 1", author: "Michael Zhang <zhan4854@umn.edu>")
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#set page("us-letter")
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#import "@preview/prooftrees:0.1.0": *
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= Homework 1
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Michael Zhang \<zhan4854\@umn.edu\>
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#let c(body) = {
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set text(gray)
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body
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}
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$#math.op("f")\(x)$
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== Problem 1
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$top : "Set"$
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$"case"_top$
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#c[Define `funsplit` in terms of `apply`.]
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$ "funsplit"(f, d) &equiv d((x) "apply"(f, x)) $
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In the case of $f = lambda b$ the following evaluation occurs:
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$ "funsplit"(lambda b, d) &equiv d((x) "apply"(lambda b, x)) \
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&equiv d((x) (b(x))) \
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&equiv d(b) $
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== Problem 2
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#c[Assuming that you have Eq sets at your disposal, use only rules (without appealing to the One True Language) to prove:
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$ plus.circle(x, 0) = x ∈ NN [x ∈ NN] $
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Please be pedantic and explicitly use a rule even for "obvious" steps. You do not have to draw a big derivation tree (which is difficult to typeset), but it should be clear which rule you are using at each step. If you are using an implicit rule that the authors did not explicitly write out, make it clear what it is. If you are not sure what a derivation tree is, whether a particular rule is allowed, or which emoji was just introduced in Emoji 15.1, use the Discord forum immediately.]
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#set math.equation(numbering: "(1)")
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First, let me reprint some rules from the textbook here to refer to them more easily from here on:
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- Reflexivity (pg. 37 in textbook)
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$ #tree(axi[$a in A$], uni[$a = a in A$]) $
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- Propositions as sets (pg. 37 in textbook)
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$ #tree(axi[$a∈A$], uni[$A "true"$]) $
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- Substitution in elements from chapter 5 (pg. 38 in textbook)
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$ #tree(
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axi[$c(x) in C(x) [x in A]$],
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axi[$a = b in A$],
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bin[$c(a) = c(b) in C(a)$],
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) $
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- Eq-introduction rule (pg. 60 in textbook)
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$ #tree(axi[$a = b in A$], uni[$"eq" in "Eq"(A, a, b)$]) $
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- Strong Eq-elimination (pg. 61 in textbook)
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$ #tree(axi[$c ∈ "Eq"(A, a, b)$], uni[$a=b∈A$]) $
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- N-introduction (pg. 63 in textbook)
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$ #tree(axi[$a in NN$], uni[$"suc"(a) in NN$]) $
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- Nat rec (pg. 64 in textbook)
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$ #tree(
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axi[$a in NN$],
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axi[$C(v) "set" [v in NN]$],
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axi[$d in C(0)$],
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axi[$e(x, y) in C("suc"(x)) [x in NN, y in C(x)]$],
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nary(4)[$"natrec"(a, d, e) in C(a)$],
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) $
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- Peano's 5th axiom (pg. 65 in textbook)
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$ #tree(
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axi[$a ∈ NN$],
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axi[$C(v) "prop" [v ∈ N]$],
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axi[$C(0) "true"$],
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axi[$C("suc"(x)) "true" [C(x) "true"]$],
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nary(4)[$C(a) "true"$]
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) $
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I can now form the derivation:
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$ #tree(
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axi[],
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uni[$NN "set"$],
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axi[$0 in NN$],
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bin(right_label: "(1)")[$0 = 0 in NN$],
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uni(right_label: "natrec")[$"natrec"(0, 0, (u, v) "suc"(v)) = 0 in NN$],
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uni(right_label: "(4)")[$"eq"_"0" in "Eq"(NN, "natrec"(0, 0, (u, v) "suc"(v)), 0)$],
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) $
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$ #tree(
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axi[$x in NN$],
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axi[$y in "Eq"(NN, "natrec"(x, 0, (u, v) "suc"(v)), x)$],
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bin(right_label: "(5)")[$"natrec"(x, 0, (u, v) "suc"(v)) = x in NN$],
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uni(right_label: "(3)")[$"suc"("natrec"(x, 0, (u, v) "suc"(v))) = "suc"(x) in NN$],
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uni(right_label: "natrec")[$"natrec"("suc"(x), 0, (u, v) "suc"(v))) = "suc"(x) in NN$],
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) $
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$ #pad(left: -4em, tree(
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axi(pad(bottom: .2em, [
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$x in NN$ \
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$"Eq"(NN, "natrec"(n, 0, (u, v) "suc"(v)), n) "set" [n in NN]$ \
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$"eq"_"0" in "Eq"(NN, "natrec"(0, 0, (u, v) "suc"(v)), 0)$ \
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$"eq"_"suc" (x, y) in "Eq"(NN, "natrec"("suc"(x), 0, (u, v) "suc"(v)), "suc"(x)) [x in NN, y in "Eq"(NN, "natrec"(x, 0, (u, v) "suc"(v)), x)]$ \
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])),
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uni(right_label: "(7, 9, 10)")[$"natrec"(x, "eq"_"0", "eq"_"suc") in "Eq"(NN, "natrec"(x, 0, (u, v) "suc"(v)), x)$],
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)) $
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#tree(
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axi[],
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uni(right_label: "(11)")[$"natrec"(x, "eq"_"0", "eq"_"suc") in "Eq"(NN, "natrec"(x, 0, (u, v) "suc"(v)), x) [x in NN]$],
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uni(right_label: "(5)")[$"natrec"(x, 0, (u, v) "suc"(v)) = x in NN [x in NN]$],
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uni(right_label: [#sym.plus.circle])[$plus.circle(x, 0) = x in NN [x in NN]$],
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)
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// #pagebreak()
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// Then, we can use this:
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// #set math.equation(numbering: none)
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// #tree(
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// axi[$0 in NN$],
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// uni(right_label: "(1)")[$0 = 0 in NN$],
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// uni(right_label: "(3)")[$"eq"_0 in "Eq"(NN, 0, 0)$],
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// )
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// Using $C(x) = NN$ and the substitution-in-elements rule, we can prove that suc distributes over equivalence:
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// #tree(
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// axi[$"suc"(x) in C(x) [x in NN]$],
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// axi[$x = y in NN$],
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// bin[$"suc"(x) = "suc"(y) in NN$],
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// )
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// Then, convert this to Eq:
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// #tree(
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// axi[$"suc"(x) = "suc"(y) in NN$],
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// axi[$x = y in NN$],
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// bin[$"eq"_"xy-suc" in "Eq"(NN, "suc"(x), "suc"(y))$],
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// )
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// We can form a lambda term over eq objects using $lambda$-introduction:
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// #tree(
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// axi[$"eq"_"xy-suc" in "Eq"(NN, "suc"(x), "suc"(y)) ["eq"_"xy" in "Eq"(NN, x, y), x in NN, y in NN]$],
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// uni[$lambda(("eq"_"xy") "eq"_"xy-suc") in "Eq"(NN, x, y) arrow.r "Eq"(NN, "suc"(x), "suc"(y))$],
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// )
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// This gives us what we need to perform induction over nats:
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// #tree(
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// axi[$"natrec"(x, "eq"_0, (u, v) "apply"(lambda(("eq"_"xy") "eq"_"xy-suc"), v)) [x in NN]$],
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// uni[]
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// )
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// #tree(
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// axi[$"i'm fucking stupid" equiv (x) "natrec"(x, 0, (u, v) "suc"(v))$],
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// uni[],
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// )
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// #tree(
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// axi[$"eq" in "Eq"(NN, "natrec"(x, 0, (u, v) "suc"(v)), x) [x in NN]$],
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// uni[$"natrec"(x, 0, (u, v) "suc"(v)) = x in NN [x in NN]$],
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// uni(right_label: "macro")[$plus.circle(x, 0) = x in NN [x in NN]$],
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// )
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// ---
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// #pagebreak()
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// #{
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// let split1 = [$"natrec" (x, 0, (u, v) "suc"(v))$]
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// tree(
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// axi[$"natrec"(x, "refl", (u, v) ("ap" v)) [x in NN]$],
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// uni[$split1 equiv x in NN [x in NN]$],
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// )
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// tree(
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// // axi[$c in "Eq"(NN, split1, 0), x)$],
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// axi[$split1 equiv x in NN [x in NN]$],
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// uni(right_label: "macro")[$plus.circle(x, 0) equiv x in NN [x in NN]$],
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// )
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// tree(
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// axi[$c in "Eq"(NN, plus.circle(x, 0), x)$],
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// uni[$plus.circle(x, 0) equiv x in NN [x in NN]$],
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// )
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// tree(
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// axi[$plus.circle(x, 0) equiv split1 [x in NN]$],
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// )
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// tree(
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// axi[$split1 equiv x [x in NN]$],
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// )
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// tree(
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// axi[$plus.circle(x, 0) equiv split1 [x in NN]$],
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// axi[$split1 equiv x [x in NN]$],
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// bin(right_label: "trans")[$plus.circle(x, 0) equiv x in NN [x in NN]$],
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// )
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// } |