completed first draft of Properties save for exercises

src/Naturals.lagda

@@ -122,7 +122,7 @@ that are on your keyboard.
 
 ## The story of creation
 
-Let's look again at the definition of natural numbers:
+Let's look again at the rules that define the natural numbers:
 
 + *Base case*: `zero` is a natural number.
 + *Inductive case*: if `m` is a natural number, then `suc m` is also a
@@ -130,7 +130,7 @@ Let's look again at the definition of natural numbers:
 
 Wait a minute! The second line defines natural numbers
 in terms of natural numbers. How can that posssibly be allowed?
-Isn't this as useless as claiming "Brexit means Brexit"?
+Isn't this as meaningless as claiming "Brexit means Brexit"?
 
 In fact, it is possible to assign our definition a meaning without
 resorting to any unpermitted circularities.  Furthermore, we can do so
@@ -142,7 +142,7 @@ no natural numbers at all.
 
     -- in the beginning, there are no natural numbers
 
-Now, we apply the two cases to all the natural numbers we know about.  The
+Now, we apply the rules to all the natural numbers we know about.  The
 base case tells us that `zero` is a natural number, so we add it to the set
 of known natural numbers.  The inductive case tells us that if `m` is a
 natural number (on the day before today) then `suc m` is also a
@@ -150,17 +150,17 @@ natural number (today).  We didn't know about any natural numbers
 before today, so the inductive case doesn't apply.
 
     -- on the first day, there is one natural number   
-    zero
+    zero : ℕ
 
 Then we repeat the process, so on the next day we know about all the
-numbers from the day before, plus any numbers added by the two cases.  The
+numbers from the day before, plus any numbers added by the rules.  The
 base case tells us that `zero` is a natural number, but we already knew
 that. But now the inductive case tells us that since `zero` was a natural
 number yesterday, then `suc zero` is a natural number today.
 
     -- on the second day, there are two natural numbers
-    zero
+    zero : ℕ
-    suc zero
+    suc zero : ℕ
 
 And we repeat the process again. Now the inductive case
 tells us that since `zero` and `suc zero` are both natural numbers, then
@@ -168,17 +168,17 @@ tells us that since `zero` and `suc zero` are both natural numbers, then
 the first of these, but the second is new.
 
     -- on the third day, there are three natural numbers
-    zero
+    zero : ℕ
-    suc zero
+    suc zero : ℕ
-    suc (suc zero)
+    suc (suc zero) : ℕ
 
 You've probably got the hang of it by now.
 
     -- on the fourth day, there are four natural numbers
-    zero
+    zero : ℕ
-    suc zero
+    suc zero : ℕ
-    suc (suc zero)
+    suc (suc zero) : ℕ
-    suc (suc (suc zero))
+    suc (suc (suc zero)) : ℕ
 
 The process continues.  On the *n*th day there will be *n* distinct
 natural numbers. Every natural number will appear on some given day.
@@ -200,14 +200,19 @@ we would have no numbers in the beginning, and still no numbers on the
 second day, and on the third, and so on.  An inductive definition lacking
 a base case is useless, as in the phrase "Brexit means Brexit".
 
-A philosopher might note that our reference to the first day, second
+
-day, and so on, implicitly involves an understanding of natural
+## Philosophy and history
+
+A philosopher might observe that our reference to the first day,
+second day, and so on, implicitly involves an understanding of natural
 numbers.  In this sense, our definition might indeed be regarded as in
-some sense circular.  We won't worry about the philosophy, but are ok
+some sense circular.  We need not let this circularity disturb us.
-with taking some intuitive notions---such as counting---as given.
+Everyone possesses a good informal understanding of the natural
+numbers, which we may take as a foundation for their formal
+description.
 
 While the natural numbers have been understood for as long as people
-could count, the inductive definition of the natural numbers is relatively
+can count, the inductive definition of the natural numbers is relatively
 recent.  It can be traced back to Richard Dedekind's paper "*Was sind
 und was sollen die Zahlen?*" (What are and what should be the
 numbers?), published in 1888, and Giuseppe Peano's book "*Arithmetices
@@ -457,52 +462,58 @@ definition to equivalent inference rules for judgements about equality.
 
 Here we assume we have already defined the infinite set of natural
 numbers, specifying the meaning of the judgment `n : ℕ`.  The first
-inference rule asserts that if `n` is a natural number then adding
+inference rule is the base case, and corresponds to line (i) of the
-zero to it gives `n`, and the second rule asserts that if adding `m`
+definition.  It asserts that if `n` is a natural number then adding
-and `n` gives `p`, then adding `suc m` and `n` gives `suc p`.
+zero to it gives `n`.  The second inference rule is the inductive
+case, and corresponds to line (ii) of the definition. It asserts that
+if adding `m` and `n` gives `p`, then adding `suc m` and `n` gives
+`suc p`.
 
-Now we can resort to a similar creation story, where now we are
+Again we resort to a creation story, where this time we are
 concerned with judgements about addition.
 
-  -- in the beginning, we know nothing about addition
+    -- in the beginning, we know nothing about addition
 
 Now, we apply the rules to all the judgment we know about.
-One rule tells us that `zero + n = n` for every natural `n`,
+The base case tells us that `zero + n = n` for every natural `n`,
-so we add all those equations.  The other rule tells us that if
+so we add all those equations.  The inductive case tells us that if
 `m + n = p` (on the day before today) then `suc m + n = suc p`
 (today).  We didn't know any equations about addition before today,
-so that rule doesn't give us any equations.
+so that rule doesn't give us any new equations.
 
-  -- on the first day, we know about addition of 0
+    -- on the first day, we know about addition of 0
-  0 + 0 = 0     0 + 1 = 1    0 + 2 = 2     ...
+    0 + 0 = 0     0 + 1 = 1    0 + 2 = 2     ...
 
 Then we repeat the process, so on the next day we know about all the
-equations from the day before, plus any equations added by the rules.  One
+equations from the day before, plus any equations added by the rules.
-rule tells us that `zero + n = n` for every natural `n`, but we already knew
+The base case tells us nothing new, but now the inductive case adds
-that. But now the other rule tells us that for every equation `m + n = p`
+more equations.
-that we knew yesterday, we know `suc m + n = suc p` today.
 
-  -- on the second day, we know about addition of 0 and 1
+    -- on the second day, we know about addition of 0 and 1
-  0 + 0 = 0     0 + 1 = 1    0 + 2 = 2     ...
+    0 + 0 = 0     0 + 1 = 1     0 + 2 = 2     0 + 3 = 3     ...
-  1 + 0 = 1     1 + 1 = 2    1 + 2 = 3     ...
+    1 + 0 = 1     1 + 1 = 2     1 + 2 = 3     1 + 3 = 4     ...
 
 And we repeat the process again.  
 
-  -- on the third day, we know about addition of 0, 1, and 2
+    -- on the third day, we know about addition of 0, 1, and 2
-  0 + 0 = 0     0 + 1 = 1    0 + 2 = 2     ...
+    0 + 0 = 0     0 + 1 = 1     0 + 2 = 2     0 + 3 = 3     ...
-  1 + 0 = 1     1 + 1 = 2    1 + 2 = 3     ...
+    1 + 0 = 1     1 + 1 = 2     1 + 2 = 3     1 + 3 = 4     ...
-  2 + 0 = 2     2 + 1 = 3    2 + 2 = 4     ...
+    2 + 0 = 2     2 + 1 = 3     2 + 2 = 4     2 + 3 = 5     ...
 
 You've probably got the hang of it by now.
 
-The process continues. On the *m*th day we will know all the equations
+    -- on the fourth day, we know about addition of 0, 1, 2, and 3
-where the first number is less than *m*.  As before, there is both a
+    0 + 0 = 0     0 + 1 = 1     0 + 2 = 2     0 + 3 = 3     ...
-base case (line (i) of the definition) and an inductive case (line
+    1 + 0 = 1     1 + 1 = 2     1 + 2 = 3     1 + 3 = 4     ...
-(ii)).
+    2 + 0 = 2     2 + 1 = 3     2 + 2 = 4     2 + 3 = 5     ...
+    3 + 0 = 3     3 + 1 = 4     3 + 2 = 5     3 + 3 = 6     ...
 
-As we can see, the reasoning that justifies inductive and
+The process continues.  On the *m*th day we will know all the
-recursive definitions is quite similar, and they might be considered
+equations where the first number is less than *m*.
-as two sides of the same coin.
+
+As we can see, the reasoning that justifies inductive and recursive
+definitions is quite similar.  They might be considered two sides of
+the same coin.
 
 
 ## Precedence
@@ -560,6 +571,11 @@ In this chapter we use the following.
     →  U+2192  RIGHTWARDS ARROW (\to, \r)
     ∸  U+2238  DOT MINUS (\.-)
 
-Each line consists of the unicode character (ℕ), the corresponding
+Each line consists of the Unicode character (ℕ), the corresponding
 code point (U+2115), the name of the character (DOUBLE-STRUCK CAPITAL N),
 and the sequence to type into Emacs to generate the character (\bN).
+
+The command \r is a little different to the others, because it gives
+access to a wide variety of Unicode arrows.  After typing \r, one can access
+the many available arrows by using the left, right, up, and down
+keys on the keyboard to navigate.

src/Properties.lagda

@@ -84,31 +84,33 @@ case*.
 
 Proofs by induction follow the structure of this definition.  To prove
 a property of natural numbers by induction, we need prove two cases.
-First is the *base case*, where we need to show the property holds for
+First is the *base case*, where we show the property holds for `zero`.
-`zero`.  Second is the *inductive case*, where we assume as the
+Second is the *inductive case*, where we assume the the property holds for
-*inductive hypothesis* that the property holds for an arbitary natural
+an arbitary natural `m` (we call this the *inductive hypothesis*), and
-`m` and then show that the property must also hold for `suc m`.
+then show that the property must also hold for `suc m`.
 
-If we write `P m` for a property of `m`, then we can write out the principle
+If we write `P m` for a property of `m`, then what we need to
-of induction as an inference rule:
+demonstrate are the following two inference rules:
 
+    ------
     P zero
-    ∀ (m : ℕ) → P m → P (suc m)
-    -----------------------------
-    ∀ (m : ℕ) → P m
 
-Let's unpack this rule.  The first hypothesis is the base case, and
+    P m
-requires we show that property `P` holds for `zero`.  The second
+    ---------
-hypothesis is the inductive case, and requires that for any natural
+    P (suc m)
-`m` assuming the the induction hypothesis `P` holds for `m` we must
+
-then show that `P` also holds for `suc m`.
+Let's unpack these rules.  The first rule is the base case, and
+requires we show that property `P` holds for `zero`.  The second rule
+is the inductive case, and requires we show that if we assume the
+inductive hypothesis, that `P` holds for `m`, then it follows that
+`P` also holds for `suc m`.
 
 Why does this work?  Again, it can be explained by a creation story.
 To start with, we know no properties.
 
     -- in the beginning, no properties are known
 
-Now, we apply the two cases to all the properties we know about.  The
+Now, we apply the two rules to all the properties we know about.  The
 base case tells us that `P zero` holds, so we add it to the set of
 known properties.  The inductive case tells us that if `P m` holds (on
 the day before today) then `P (suc m)` also holds (today).  We didn't
@@ -119,8 +121,8 @@ apply.
     P zero
 
 Then we repeat the process, so on the next day we know about all the
-properties from the day before, plus any properties added by the two
+properties from the day before, plus any properties added by the rules.
-cases.  The base case tells us that `P zero` holds, but we already
+The base case tells us that `P zero` holds, but we already
 knew that. But now the inductive case tells us that since `P zero`
 held yesterday, then `P (suc zero)` holds today.
 
@@ -157,14 +159,18 @@ In order to prove associativity, we take `P m` to be the property
 
     (m + n) + p ≡ m + (n + p)
 
-Then the appropriate instance of the inference rule becomes:
+Then the appropriate instances of the inference rules, which
+we must show to hold, become:
 
+    n : ℕ     p : ℕ
+    -------------------------------
     (zero + n) + p ≡ zero + (n + p)
-    ∀ (m : ℕ) → (m + n) + p ≡ m + (n + p)  →  (suc m + n) + p ≡ (suc m + n) + p
-    -----------------------------------------------------------------------------
-    ∀ (m : ℕ) → (m + n) + p ≡ m + (n + p)
 
-In the inference rule, `n` and `p` are any arbitary natural numbers, so when we
+    (m + n) + p ≡ m + (n + p)
+    ---------------------------------
+    (suc m + n) + p ≡ (suc m + n) + p
+
+In the inference rules, `n` and `p` are any arbitary natural numbers, so when we
 are done with the proof we know it holds for any `n` and `p` as well as any `m`.
 
 Recall the definition of addition.
@@ -221,11 +227,11 @@ induction hypothesis.
 
 ## Encoding the proof in Agda
 
-We can encode this proof in Agda as follows.
+We encode this proof in Agda as follows.
 \begin{code}
-assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+-- assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
-assoc+ zero n p = refl
+-- assoc+ zero n p = refl
-assoc+ (suc m) n p rewrite assoc+ m n p = refl
+-- assoc+ (suc m) n p rewrite assoc+ m n p = refl
 \end{code}
 Here we have named the proof `assoc+`.  In Agda, identifiers can consist of
 any sequence of characters not including spaces or the characters `@.(){};_`.
@@ -236,13 +242,15 @@ proposition
 
     ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
 
-Such a proof is a function that takes three natural
+The upside down A is pronounced "for all", and the proposition
-numbers---corresponding to `m`, `n`, and `p`---and returns
+asserts that for all natural numbers `m`, `n`, and `p` that
-a proof of the proposition `(m + n) + p ≡ m + (n + p)`.
+the equations `(m + n) + p ≡ m + (n + p)` holds.  Such a proof
-Proof by induction corresponds exactly to a recursive
+is a function that accepts three natural numbers---corresponding to
-definition.  Here we are inducting on the first
+to `m`, `n`, and `p`---and returns a proof of the equation.
-argument `m`, and leaving the other two arguments `n` and `p`
+
-fixed.
+Proof by induction corresponds exactly to a recursive definition.
+Here we are induct on the first argument `m`, and leave the other two
+arguments `n` and `p` fixed.
 
 The base case corresponds to instantiating `m` by
 pattern matching on `zero`, so what we are required to
@@ -269,18 +277,303 @@ becomes:
 
 After rewriting with the inductive hypothesis these two terms are
 equal, and the proof is again given by `refl`.
-
 Rewriting by a given equation is indicated by the keyword `rewrite`
-followed by a proof of that equation.  Here the inductive hypothesis
+followed by a proof of that equation.
-is not assumed, but instead proved by a recursive invocation of the
-function we are definining, `assoc+ m n p`.  As with addition, this is
-well founded because associativity of larger numbers is defined in
-terms of associativity of smaller numbers.  In this case,
-`assoc (suc m) n p` is defined in terms of `assoc m n p`.
 
-This correspondence between proof by induction and definition by
+Here the inductive hypothesis is not assumed, but instead proved by a
+recursive invocation of the function we are definining, `assoc+ m n p`.
+As with addition, this is well founded because associativity of
+larger numbers is proved in terms of associativity of smaller numbers.
+In this case, `assoc (suc m) n p` is proved using `assoc m n p`.
+
+The correspondence between proof by induction and definition by
 recursion is one of the most appealing aspects of Agda.
 
+
+## Building proofs interactively
+
+Agda is designed to be used with the Emacs text editor, and the two
+in combination provide features that help to create proofs interactively.
+
+Begin by typing
+
+    assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+    assoc+ m n p = ?
+
+The question mark indicates that you would like Agda to help with
+filling in that part of the code.  If you type `^C ^L` (control-C
+followed by control-L), the question mark will be replaced.
+
+    assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+    assoc+ m n p = { }0
+
+The empty braces are called a *hole*, and 0 is a number used for
+referring to the hole.  The hole may display highlighted in green.
+Emacs will also create a new window at the bottom of the screen
+displaying the text
+
+    ?0 : ((m + n) + p) ≡ (m + (n + p))
+
+This indicates that hole 0 is to be filled in with a proof of
+the stated judgement.
+
+We wish to prove the proposition by induction on `m`.  More
+the cursor into the hole and type `^C ^C`.  You will be given
+the prompt:
+
+    pattern variables to case:
+
+Typing `m` will cause a split on that variable, resulting
+in an update to the code.
+
+    assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+    assoc+ zero n p = { }0
+    assoc+ (suc m) n p = { }1
+
+There are now two holes, and the window at the bottom tells you what
+each is required to prove:
+
+    ?0 : ((zero + n) + p) ≡ (zero + (n + p))
+    ?1 : ((suc m + n) + p) ≡ (suc m + (n + p))
+
+Going into hole 0 and typing `^C ^,` will display the text:
+
+    Goal: (n + p) ≡ (n + p)
+    ————————————————————————————————————————————————————————————
+    p : ℕ
+    n : ℕ
+
+This indicates that after simplification the goal for hole 0 is as
+stated, and that variables `p` and `n` of the stated types are
+available to use in the proof.  The proof of the given goal is
+trivial, and going into the goal and typing `^C ^R` will fill it in,
+renumbering the remaining hole to 0:
+
+    assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+    assoc+ zero n p = refl
+    assoc+ (suc m) n p = { }0
+
+Going into the new hold 0 and typing `^C ^,` will display the text:
+
+    Goal: suc ((m + n) + p) ≡ suc (m + (n + p))
+    ————————————————————————————————————————————————————————————
+    p : ℕ
+    n : ℕ
+    m : ℕ
+
+Again, this gives the simplified goal and the available variables.
+In this case, we need to rewrite by the induction
+hypothesis, so let's edit the text accordingly:
+
+    assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+    assoc+ zero n p = refl
+    assoc+ (suc m) n p rewrite assoc+ m n p = { }0
+
+Going into the remaining hole and typing `^C ^,` will show the
+goal is now trivial:
+
+    Goal: suc (m + (n + p)) ≡ suc (m + (n + p))
+    ————————————————————————————————————————————————————————————
+    p : ℕ
+    n : ℕ
+    m : ℕ
+
+The proof of the given goal is trivial, and going into the goal and
+typing `^C ^R` will fill it in, completing the proof:
+
+    assoc+ : ∀ (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+    assoc+ zero n p = refl
+    assoc+ (suc m) n p rewrite assoc+ m n p = refl
+
+
+## Creation, one last time
+
+Again, it may be helpful to view the recursive definition as
+a creation story.  This time we are concerned with judgements
+asserting associativity.
+
+     -- in the beginning, we know nothing about associativity
+
+Now, we apply the rules to all the judgements we know about.  The base
+case tells us that `(zero + n) + p ≡ zero + (n + p)` for every natural
+`n` and `p`.  The inductive case tells us that if `(m + n) + p ≡ m +
+(n + p)` (on the day before today) then `(suc m + n) + p ≡ suc m + (n
++ p)` (today).  We didn't know any judgments about associativity
+before today, so that rule doesn't give us any new judgments.
+
+    -- on the first day, we know about associativity of 0
+    (0 + 0) + 0 ≡ 0 + (0 + 0)   ...   (0 + 4) + 5 ≡ 0 + (4 + 5)   ...
+    
+Then we repeat the process, so on the next day we know about all the
+judgements from the day before, plus any judgements added by the rules.
+The base case tells us nothing new, but now the inductive case adds
+more judgements.
+
+    -- on the second day, we know about associativity of 0 and 1
+    (0 + 0) + 0 ≡ 0 + (0 + 0)   ...   (0 + 4) + 5 ≡ 0 + (4 + 5)   ...
+    (1 + 0) + 0 ≡ 1 + (0 + 0)   ...   (1 + 4) + 5 ≡ 1 + (4 + 5)   ...
+
+And we repeat the process again.
+
+    -- on the third day, we know about associativity of 0, 1, and 2
+    (0 + 0) + 0 ≡ 0 + (0 + 0)   ...   (0 + 4) + 5 ≡ 0 + (4 + 5)   ...
+    (1 + 0) + 0 ≡ 1 + (0 + 0)   ...   (1 + 4) + 5 ≡ 1 + (4 + 5)   ...
+    (2 + 0) + 0 ≡ 2 + (0 + 0)   ...   (2 + 4) + 5 ≡ 2 + (4 + 5)   ...
+
+You've probably got the hang of it by now.
+
+    -- on the fourth day, we know about associativity of 0, 1, 2, and 3
+    (0 + 0) + 0 ≡ 0 + (0 + 0)   ...   (0 + 4) + 5 ≡ 0 + (4 + 5)   ...
+    (1 + 0) + 0 ≡ 1 + (0 + 0)   ...   (1 + 4) + 5 ≡ 1 + (4 + 5)   ...
+    (2 + 0) + 0 ≡ 2 + (0 + 0)   ...   (2 + 4) + 5 ≡ 2 + (4 + 5)   ...
+
+The process continues.  On the *m*th day we will know all the
+judgements where the first number is less than *m*.
+
+
+## Our second proof: commutativity
+
+Another important property of addition is that it is commutative.  To show
+this by induction, we take `P m` to be the property
+
+    m + n ≡ n + m
+
+Then the appropriate instances of the inference rules, which
+we must show to hold, become:
+
+    n : ℕ
+    -------------------
+    zero + n ≡ n + zero
+
+    m + n ≡ n + m
+    ---------------------
+    suc m + n ≡ n + suc m
+
+In the inference rules, `n` is any arbitary natural number, so when we
+are done with the proof we know it holds for any `n` as well as any `m`.
+
+By equation (i) of the definition of addition, the left-hand side of
+the conclusion of the base case simplifies to `n`, so the base case
+will hold if we can show
+
+    n + zero ≡ n    -- (x)
+
+By equation (ii) of the definition of addition, the left-hand side of
+the conclusion of the inductive case simplifies to `suc (m + n)`, so
+the inductive case will hold if we can show
+
+    n + suc m ≡ suc (n + m)    -- (xi)
+
+and then apply the inductive hypothesis to rewrite `m + n` to `n + m`.
+We use induction two more times, on `n` this time rather than `m`, to
+show both (x) and (xi).
+
+Here is the proof written out in full, using tabular form.
+
+Proposition.
+
+    m + n ≡ n + m
+
+Proof. By induction on `m`.
+
+Base case.
+
+       zero + n
+    ≡    (i)
+       n
+    ≡    (x)
+       n + zero
+
+Inductive case.
+
+       suc m + n
+    ≡    (ii)
+       suc (m + n)
+    ≡    (inductive hypothesis)
+       suc (n + m)
+    ≡    (xi)
+       n + suc m
+       
+QED.
+
+As with other tabular prooofs, it is best understood by reading from top
+down and bottom up and meeting in the middle.
+
+We now prove each of the two required lemmas, (x) and (xi).
+
+Lemma (x).
+
+    n + zero ≡ n
+
+Proof. By induction on `n`.
+
+Base case.
+
+       zero + zero
+    ≡    (i)
+       zero
+
+Inductive case.
+
+       suc n + zero
+    ≡    (ii)
+       suc (n + zero)
+    ≡    (inductive hypothesis)
+       suc n
+
+QED.
+
+Lemma (xi).
+
+    n + suc m ≡ suc (n + m)
+
+Proof. By induction on `n`.
+
+Base case.
+
+       zero + suc m
+    ≡    (i)
+       suc m
+    ≡    (i)
+       suc (zero + m)
+
+Inductive case.
+
+       suc n + suc m
+    ≡    (ii)
+       suc (n + suc m)
+    ≡    (inductive hypothesis)
+       suc (suc (n + m))
+    ≡    (ii)
+       suc (suc n + m)
+
+QED.
+
+## Encoding the proofs in Agda
+
+These proofs can be encoded concisely in Agda.
+\begin{code}
+com+zero : ∀ (n : ℕ) → n + zero ≡ n
+com+zero zero = refl
+com+zero (suc n) rewrite com+zero n = refl
+
+com+suc : ∀ (m n : ℕ) → n + suc m ≡ suc (n + m)
+com+suc m zero = refl
+com+suc m (suc n) rewrite com+suc m n = refl
+
+com : ∀ (m n : ℕ) → m + n ≡ n + m
+com zero n rewrite com+zero n = refl
+com (suc m) n rewrite com+suc m n | com m n = refl
+\end{code}
+Here we have renamed Lemma (x) and (xi) to `com+zero` and `com+suc`,
+respectively.  In the final line, rewriting with two equations
+(lemma (xi) and the inductive hypothesis) is indicated by
+separating the two proofs of the relevant equations by a vertical bar;
+the rewrites on the left is performed before that on the right.
+
+
+
+
 ## Unicode
 
 In this chapter we use the following unicode.