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theory Nat
imports Identity
begin
axiomatization
Nat :: \<open>o\<close> and
zero :: \<open>o\<close> ("0") and
suc :: \<open>o \<Rightarrow> o\<close> and
NatInd :: \<open>(o \<Rightarrow> o) \<Rightarrow> o \<Rightarrow> (o \<Rightarrow> o \<Rightarrow> o) \<Rightarrow> o \<Rightarrow> o\<close>
where
NatF: "Nat: U i" and
Nat_zero: "0: Nat" and
Nat_suc: "n: Nat \<Longrightarrow> suc n: Nat" and
NatE: "\<lbrakk>
n: Nat;
c\<^sub>0: C 0;
\<And>k rec. \<lbrakk>k: Nat; rec: C k\<rbrakk> \<Longrightarrow> f k rec: C (suc k);
\<And>n. n: Nat \<Longrightarrow> C n: U i
\<rbrakk> \<Longrightarrow> NatInd (fn n. C n) c\<^sub>0 (fn k rec. f k rec) n: C n" and
Nat_comp_zero: "\<lbrakk>
c\<^sub>0: C 0;
\<And>k rec. \<lbrakk>k: Nat; rec: C k\<rbrakk> \<Longrightarrow> f k rec: C (suc k);
\<And>n. n: Nat \<Longrightarrow> C n: U i
\<rbrakk> \<Longrightarrow> NatInd (fn n. C n) c\<^sub>0 (fn k rec. f k rec) 0 \<equiv> c\<^sub>0" and
Nat_comp_suc: "\<lbrakk>
n: Nat;
c\<^sub>0: C 0;
\<And>k rec. \<lbrakk>k: Nat; rec: C k\<rbrakk> \<Longrightarrow> f k rec: C (suc k);
\<And>n. n: Nat \<Longrightarrow> C n: U i
\<rbrakk> \<Longrightarrow>
NatInd (fn n. C n) c\<^sub>0 (fn k rec. f k rec) (suc n) \<equiv>
f n (NatInd (fn n. C n) c\<^sub>0 (fn k rec. f k rec) n)"
lemmas
[form] = NatF and
[intr, intro] = Nat_zero Nat_suc and
[elim "?n"] = NatE and
[comp] = Nat_comp_zero Nat_comp_suc
abbreviation "NatRec C \<equiv> NatInd (fn _. C)"
abbreviation one ("1") where "1 \<equiv> suc 0"
abbreviation two ("2") where "2 \<equiv> suc 1"
abbreviation three ("3") where "3 \<equiv> suc 2"
abbreviation four ("4") where "4 \<equiv> suc 3"
abbreviation five ("5") where "5 \<equiv> suc 4"
abbreviation six ("6") where "6 \<equiv> suc 5"
abbreviation seven ("7") where "7 \<equiv> suc 6"
abbreviation eight ("8") where "8 \<equiv> suc 7"
abbreviation nine ("9") where "9 \<equiv> suc 8"
section \<open>Basic arithmetic\<close>
subsection \<open>Addition\<close>
definition add (infixl "+" 120) where "m + n \<equiv> NatRec Nat m (K suc) n"
Lemma add_type [type]:
assumes "m: Nat" "n: Nat"
shows "m + n: Nat"
unfolding add_def by typechk
Lemma add_zero [comp]:
assumes "m: Nat"
shows "m + 0 \<equiv> m"
unfolding add_def by compute
Lemma add_suc [comp]:
assumes "m: Nat" "n: Nat"
shows "m + suc n \<equiv> suc (m + n)"
unfolding add_def by compute
Lemma (def) zero_add:
assumes "n: Nat"
shows "0 + n = n"
apply (elim n)
\<^item> by (compute; intro)
\<^item> vars _ ih by compute (eq ih; refl)
done
Lemma (def) suc_add:
assumes "m: Nat" "n: Nat"
shows "suc m + n = suc (m + n)"
apply (elim n)
\<^item> by compute refl
\<^item> vars _ ih by compute (eq ih; refl)
done
Lemma (def) suc_eq:
assumes "m: Nat" "n: Nat"
shows "p: m = n \<Longrightarrow> suc m = suc n"
by (eq p) intro
Lemma (def) add_assoc:
assumes "l: Nat" "m: Nat" "n: Nat"
shows "l + (m + n) = l + m+ n"
apply (elim n)
\<^item> by compute intro
\<^item> vars _ ih by compute (eq ih; refl)
done
Lemma (def) add_comm:
assumes "m: Nat" "n: Nat"
shows "m + n = n + m"
apply (elim n)
\<^item> by (compute; rule zero_add[symmetric])
\<^item> vars n ih
proof compute
have "suc (m + n) = suc (n + m)" by (eq ih) refl
also have ".. = suc n + m" by (rewr eq: suc_add) refl
finally show "?" by this
qed
done
subsection \<open>Multiplication\<close>
definition mul (infixl "*" 121) where "m * n \<equiv> NatRec Nat 0 (K $ add m) n"
Lemma mul_type [type]:
assumes "m: Nat" "n: Nat"
shows "m * n: Nat"
unfolding mul_def by typechk
Lemma mul_zero [comp]:
assumes "n: Nat"
shows "n * 0 \<equiv> 0"
unfolding mul_def by compute
Lemma mul_one [comp]:
assumes "n: Nat"
shows "n * 1 \<equiv> n"
unfolding mul_def by compute
Lemma mul_suc [comp]:
assumes "m: Nat" "n: Nat"
shows "m * suc n \<equiv> m + m * n"
unfolding mul_def by compute
Lemma (def) zero_mul:
assumes "n: Nat"
shows "0 * n = 0"
apply (elim n)
\<^item> by compute refl
\<^item> vars n ih
proof compute
have "0 + 0 * n = 0 + 0 " by (eq ih) refl
also have ".. = 0" by compute refl
finally show "?" by this
qed
done
Lemma (def) suc_mul:
assumes "m: Nat" "n: Nat"
shows "suc m * n = m * n + n"
apply (elim n)
\<^item> by compute refl
\<^item> vars n ih
proof (compute, rewr eq: \<open>ih:_\<close>)
have "suc m + (m * n + n) = suc (m + ?)" by (rule suc_add)
also have ".. = suc (m + m * n + n)" by (rewr eq: add_assoc) refl
finally show "?" by this
qed
done
Lemma (def) mul_dist_add:
assumes "l: Nat" "m: Nat" "n: Nat"
shows "l * (m + n) = l * m + l * n"
apply (elim n)
\<^item> by compute refl
\<^item> vars n ih
proof compute
have "l + l * (m + n) = l + (l * m + l * n)" by (eq ih) refl
also have ".. = l + l * m + l * n" by (rule add_assoc)
also have ".. = l * m + l + l * n" by (rewr eq: add_comm) refl
also have ".. = l * m + (l + l * n)" by (rewr eq: add_assoc) refl
finally show "?" by this
qed
done
Lemma (def) mul_assoc:
assumes "l: Nat" "m: Nat" "n: Nat"
shows "l * (m * n) = l * m * n"
apply (elim n)
\<^item> by compute refl
\<^item> vars n ih
proof compute
have "l * (m + m * n) = l * m + l * (m * n)" by (rule mul_dist_add)
also have ".. = l * m + l * m * n" by (rewr eq: \<open>ih:_\<close>) refl
finally show "?" by this
qed
done
Lemma (def) mul_comm:
assumes "m: Nat" "n: Nat"
shows "m * n = n * m"
apply (elim n)
\<^item> by compute (rewr eq: zero_mul, refl)
\<^item> vars n ih
proof (compute, rule pathinv)
have "suc n * m = n * m + m" by (rule suc_mul)
also have ".. = m + m * n"
by (rewr eq: \<open>ih:_\<close>, rewr eq: add_comm) refl
finally show "?" by this
qed
done
end
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