added Relations
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@ -456,6 +456,9 @@ definition to equivalent inference rules for judgements about equality.
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------------
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------------
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zero + n = n
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zero + n = n
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m : ℕ
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n : ℕ
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p : ℕ
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m + n = p
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m + n = p
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-------------------
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-------------------
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(suc m) + n = suc p
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(suc m) + n = suc p
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@ -21,6 +21,7 @@ everything in the previous chapter is also found in the library module
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We also require propositional equality. -->
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We also require propositional equality. -->
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\begin{code}
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\begin{code}
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open import Naturals using (ℕ; zero; suc; _+_; _*_; _∸_)
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open import Relation.Binary.PropositionalEquality using (_≡_; refl)
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open import Relation.Binary.PropositionalEquality using (_≡_; refl)
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\end{code}
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\end{code}
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@ -72,15 +73,9 @@ The answer is yes! We can prove a property holds for all naturals using
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## Proof by induction
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## Proof by induction
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Recall the definition of natural numbers.
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Recall the definition of natural numbers consists of a *base case*
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\begin{code}
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which tells us that `zero` is a natural, and an *inductive case*
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data ℕ : Set where
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which tells us that if `m` is a natural then `suc m` is also a natural.
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zero : ℕ
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suc : ℕ → ℕ
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\end{code}
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This tells us that `zero` is a natural---the *base case*---and that if
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`m` is a natural then `suc m` is also a natural---the *inductive
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case*.
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Proofs by induction follow the structure of this definition. To prove
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Proofs by induction follow the structure of this definition. To prove
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a property of natural numbers by induction, we need prove two cases.
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a property of natural numbers by induction, we need prove two cases.
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@ -173,12 +168,10 @@ we must show to hold, become:
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In the inference rules, `n` and `p` are any arbitary natural numbers, so when we
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In the inference rules, `n` and `p` are any arbitary natural numbers, so when we
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are done with the proof we know it holds for any `n` and `p` as well as any `m`.
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are done with the proof we know it holds for any `n` and `p` as well as any `m`.
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Recall the definition of addition.
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Recall the definition of addition has two clauses.
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\begin{code}
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_+_ : ℕ → ℕ → ℕ
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zero + n = n -- (i)
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zero + n = n -- (i)
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(suc m) + n = suc (m + n) -- (ii)
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(suc m) + n = suc (m + n) -- (ii)
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\end{code}
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For the base case, we must show:
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For the base case, we must show:
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@ -561,10 +554,9 @@ com+suc : ∀ (m n : ℕ) → n + suc m ≡ suc (n + m)
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com+suc m zero = refl
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com+suc m zero = refl
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com+suc m (suc n) rewrite com+suc m n = refl
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com+suc m (suc n) rewrite com+suc m n = refl
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com : ∀ (m n : ℕ) → m + n ≡ n + m
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com+ : ∀ (m n : ℕ) → m + n ≡ n + m
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com zero n rewrite com+zero n = refl
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com+ zero n rewrite com+zero n = refl
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com (suc m) n rewrite com+suc m n | com m n = refl
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com+ (suc m) n rewrite com+suc m n | com+ m n = refl
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\end{code}
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\end{code}
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Here we have renamed Lemma (x) and (xi) to `com+zero` and `com+suc`,
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Here we have renamed Lemma (x) and (xi) to `com+zero` and `com+suc`,
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respectively. In the final line, rewriting with two equations
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respectively. In the final line, rewriting with two equations
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158
src/Relations.lagda
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158
src/Relations.lagda
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@ -0,0 +1,158 @@
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---
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title : "Relations: Inductive Definition of Relations"
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layout : page
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permalink : /Relations
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---
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After having defined operations such as addition and multiplication,
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the next step is to define relations, such as *less than or equal*.
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## Imports
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\begin{code}
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open import Naturals using (ℕ; zero; suc; _+_; _*_; _∸_)
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open import Properties using (com+; com+zero; com+suc)
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open import Relation.Binary.PropositionalEquality using (_≡_; refl)
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\end{code}
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## Defining relations
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The relation *less than or equal* has an infinite number of
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instances. Here are just a few of them:
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0 ≤ 0 0 ≤ 1 0 ≤ 2 0 ≤ 3 ...
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1 ≤ 1 1 ≤ 2 1 ≤ 3 ...
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2 ≤ 2 2 ≤ 3 ...
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3 ≤ 3 ...
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...
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And yet, we can write a finite definition that encompasses
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all of these instances in just a few lines. Here is the
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definition as a pair of inference rules:
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z≤n --------
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zero ≤ n
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m ≤ n
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s≤s -------------
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suc m ≤ suc n
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And here is the definition in Agda:
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\begin{code}
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data _≤_ : ℕ → ℕ → Set where
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z≤n : ∀ {m : ℕ} → zero ≤ m
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s≤s : ∀ {m n : ℕ} → m ≤ n → suc m ≤ suc n
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\end{code}
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Here `z≤n` and `s≤s` (with no spaces) are constructor names,
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while `m ≤ m`, and `m ≤ n` and `suc m ≤ suc n` (with spaces)
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are propositions.
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Both definitions above tell us the same two things:
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+ *Base case*: for all naturals `n`, the proposition `zero ≤ n` holds
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+ *Inductive case*: for all naturals `m` and `n`, if the proposition
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`m ≤ n` holds, then the proposition `suc m ≤ suc n` holds.
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In fact, they each give us a bit more detail:
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+ *Base case*: for all naturals `n`, the constructor `z≤n`
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produces evidence that `zero ≤ n` holds.
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+ *Inductive case*: for all naturals `m` and `n`, the constructor
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`s≤s` takes evidence that `m ≤ n` holds into evidence that
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`suc m ≤ suc n` holds.
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Here we have used the word *evidence* as interchangeable with the
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word *proof*. We will tend to say *evidence* when we want to stress
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that proofs are just terms in Agda.
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For example, here in inference rule notation is the proof that
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`2 ≤ 4`.
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z≤n -----
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0 ≤ 2
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s≤s -------
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1 ≤ 3
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s≤s ---------
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2 ≤ 4
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And here is the corresponding Agda proof.
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\begin{code}
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ex₁ : 2 ≤ 4
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ex₁ = s≤s (s≤s z≤n)
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\end{code}
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## Implicit arguments
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In the Agda definition, the two lines defining the constructors
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use `∀`, very similar to our use of `∀` in propositions such as:
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com+ : ∀ (m n : ℕ) → m + n ≡ n + m
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However, here the declarations are surrounded by curly braces `{ }`
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rather than parentheses `( )`. This means that the arguments are
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*implicit* and need not be written explicitly. Thus, we would write,
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for instance, `com+ m n` for the proof that `m + n ≡ n + m`, but
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here will write `z≤n` for the proof that `m ≤ m`, leaving the `m` implicit,
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or if `m≤n` is evidence that `m ≤ n`, we write write `s≤s m≤n` for the
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evidence that `suc m ≤ suc n`, leaving both `m` and `n` implicit.
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It is possible to provide implicit arguments explicitly if we wish.
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For instance, here is the Agda proof that `2 ≤ 4` repeated, with the
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implicit arguments made explicit.
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\begin{code}
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ex₂ : 2 ≤ 4
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ex₂ = s≤s {1} {3} (s≤s {0} {2} (z≤n {2}))
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\end{code}
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## Proving properties of inductive definitions.
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\begin{code}
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infix 4 _≤_
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refl≤ : ∀ (n : ℕ) → n ≤ n
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refl≤ zero = z≤n
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refl≤ (suc n) = s≤s (refl≤ n)
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trans≤ : ∀ {m n p : ℕ} → m ≤ n → n ≤ p → m ≤ p
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trans≤ z≤n n≤p = z≤n
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trans≤ (s≤s m≤n) (s≤s n≤p) = s≤s (trans≤ m≤n n≤p)
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antisym≤ : ∀ {m n : ℕ} → m ≤ n → n ≤ m → m ≡ n
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antisym≤ z≤n z≤n = refl
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antisym≤ (s≤s m≤n) (s≤s n≤m) rewrite antisym≤ m≤n n≤m = refl
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mono+≤ : ∀ (m p q : ℕ) → p ≤ q → m + p ≤ m + q
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mono+≤ zero p q p≤q = p≤q
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mono+≤ (suc m) p q p≤q = s≤s (mono+≤ m p q p≤q)
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mono+≤′ : ∀ (m n p : ℕ) → m ≤ n → m + p ≤ n + p
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mono+≤′ m n p m≤n rewrite com+ m p | com+ n p = mono+≤ p m n m≤n
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mono+≤″ : ∀ (m n p q : ℕ) → m ≤ n → p ≤ q → m + p ≤ n + q
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mono+≤″ m n p q m≤n p≤q = trans≤ (mono+≤′ m n p m≤n) (mono+≤ n p q p≤q)
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inverse+∸≤ : ∀ (m n : ℕ) → m ≤ n → (n ∸ m) + m ≡ n
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inverse+∸≤ zero n z≤n rewrite com+zero n = refl
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inverse+∸≤ (suc m) (suc n) (s≤s m≤n)
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rewrite com+suc m (n ∸ m) | inverse+∸≤ m n m≤n = refl
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data _<_ : ℕ → ℕ → Set where
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z<s : ∀ {n : ℕ} → zero < suc n
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s<s : ∀ {m n : ℕ} → m < n → suc m < suc n
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infix 4 _<_
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<implies≤ : ∀ (m n : ℕ) → m < n → suc m ≤ n
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<implies≤ m n m<n = ?
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≤implies< : ∀ (m n : ℕ) → suc m ≤ n → m < n
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≤implies< m n m≤n = ?
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\end{code}
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## Unicode
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In this chapter we use the following unicode.
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≡ U+2261 IDENTICAL TO (\==)
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∀ U+2200 FOR ALL (\forall)
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λ U+03BB GREEK SMALL LETTER LAMBDA (\Gl, \lambda)
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