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Pierre Letouzey
proglmfi
Commits
8a7746f6
Commit
8a7746f6
authored
Nov 23, 2020
by
Pierre Letouzey
Browse files
seance4.v translated in seance4english.v (and with slightly extended comments)
parent
157fd6a8
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1
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seance4english.v
0 → 100644
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8a7746f6
(
**
*
An
introduction
to
dependent
types
*
)
(
**
M2
LMFI
*
)
(
**
NB
:
in
all
comments
below
,
the
enclosing
[
]
are
just
a
way
to
cite
a
fragment
of
Coq
code
.
See
the
documentation
of
the
coqdoc
tool
for
more
:
https:
//coq.inria.fr/distrib/current/refman/using/tools/coqdoc.html
*
)
(
**
Programming
with
dependent
type
is
one
of
the
key
feature
of
Coq
.
Such
programming
is
also
possible
in
other
languages
such
as
Haskell
and
OCaml
(
see
the
GADT
types
,
for
Generalized
Abstract
DataTypes
).
But
here
the
ability
to
mix
types
and
values
can
be
very
powerful
.
But
first
,
what
is
a
dependent
type
?
*
)
(
**
**
A
first
example
:
tuples
*
)
(
**
Reminder
:
the
type
of
pairs
*
)
(
**
That
'
s
a
type
named
[
prod
],
with
syntaxe
[
*
]
for
it
(
in
scope
delimited
if
necessary
by
[
%
type
]).
The
constructor
is
named
[
pair
],
syntaxe
[(
,
)].
*
)
Check
(
1
,
2
).
Check
pair
1
2.
(
*
same
as
[(
1
,
2
)]
*
)
Print
prod
.
Check
(
prod
nat
nat
).
Check
(
nat
*
nat
)
%
type
.
(
*
same
as
[(
prod
nat
nat
)]
*
)
Check
(
nat
>
nat
*
nat
).
(
*
no
[
%
type
]
needed
when
a
type
is
expected
*
)
(
**
For
triples
and
other
n

tuples
,
Coq
provides
some
"syntactic sugar"
:
[
nat
*
nat
*
nat
]
is
synonym
to
[(
nat
*
nat
)
*
nat
]
and
[(
1
,
2
,
3
)]
is
actually
[((
1
,
2
),
3
)].
*
)
Check
(
1
,
2
,
3
).
Check
((
1
,
2
),
3
).
(
*
the
same
...
*
)
Unset
Printing
Notations
.
Check
(
1
,
2
,
3
).
Set
Printing
Notations
.
(
**
How
to
write
the
type
of
n

uples
for
a
given
n
?
*
)
(
**
First
try
,
via
an
inductive
type
*
)
Inductive
nuple_ind
(
A
:
Type
)
:=

Nothing

More
:
A
>
nuple_ind
A
>
nuple_ind
A
.
Arguments
Nothing
{
A
}
.
(
*
Makes
1
st
argument
of
[
Nothing
]
be
implicit
*
)
Arguments
More
{
A
}
.
(
*
Same
for
[
More
]
*
)
Check
(
More
1
(
More
2
(
More
3
Nothing
))).
(
**
Actually
,
[
nuple_ind
]
is
just
a
clone
of
the
types
of
listes
...
And
we
cannot
speak
easily
of
a
particular
size
(
e
.
g
.
triple
),
since
the
type
system
doesn
'
t
distinguish
a
couple
for
a
triple
.
*
)
Check
(
More
1
(
More
2
Nothing
)).
(
*
:
nuple_ind
nat
*
)
(
**
Better
:
let
'
s
"program"
the
desired
type
,
starting
from
the
number
[
n
].
The
obtained
type
depends
from
the
value
of
this
number
:
that
'
s
hence
a
dependent
type
.
*
)
Fixpoint
nuple
(
A
:
Type
)
n
:=
match
n
with

0
=>
A

S
n
=>
((
nuple
A
n
)
*
A
)
%
type
end
.
(
**
Nota
:
here
the
[
%
type
]
above
is
mandatory
otherwise
Coq
interprets
[
*
]
as
a
multiplication
by
default
,
instead
of
the
pair
of
types
.
*
)
Locate
"*"
.
Check
nuple
.
(
*
Type
>
nat
>
Type
*
)
Compute
(
nuple
nat
0
).
(
*
Synonym
for
nat
*
)
Compute
(
nuple
nat
1
).
(
*
Synonym
for
nat
*
nat
,
e
.
g
.
the
type
of
couples
*
)
Compute
(
nuple
nat
5
).
(
*
Sextuples
of
numbers
,
beware
of
the
"ofbyone"
*
)
(
**
And
we
could
even
reuse
the
earlier
"syntactic sugar"
of
Coq
n

uples
for
building
examples
.
*
)
Check
(
1
,
2
,
3
,
4
,
5
,
6
)
:
nuple
nat
5.
(
**
More
generally
,
we
could
also
"program"
some
n

uple
examples
:
*
)
Fixpoint
ints
n
:
nuple
nat
n
:=
match
n
with

0
=>
0

S
n
'
=>
((
ints
n
'
),
n
)
end
.
(
**
Note
that
in
the
previous
[
match
]
the
two
branches
have
different
types
:
[
nat
]
in
the
first
case
and
[(
nuple
nat
n
'
*
nat
)]
in
the
second
case
.
This
is
a
dependent
match
,
whose
typing
is
quite
different
from
all
earlier
[
match
]
we
'
ve
done
,
where
all
branches
were
having
the
same
common
type
.
Here
Coq
is
clever
enough
to
notice
that
these
different
types
are
actually
the
correct
instances
of
the
claimed
type
[
nuple
nat
n
],
respectively
[
nuple
nat
0
]
and
[
nuple
nat
(
S
n
'
)].
But
Coq
would
not
be
clever
enough
to
guess
the
output
type
[
nuple
nat
n
]
by
itself
,
here
it
is
mandatory
to
write
it
.
*
)
Compute
ints
5.
Compute
ints
99
:
nuple
nat
99.
(
*
a
100

uple
*
)
(
**
A
good
way
to
notice
a
function
over
a
dependent
type
:
There
'
s
a
[
forall
]
in
its
type
:
*
)
Check
ints
.
(
*
forall
n
,
nuple
nat
n
*
)
(
**
Indeed
,
no
way
to
write
this
type
with
the
non

dependent
arrow
[
nat
>
nuple
nat
???
],
since
we
need
here
to
name
the
left
argument
which
is
used
at
the
end
.
*
)
(
**
We
'
ll
see
below
that
another
solution
is
possible
to
represent
n

uples
:
the
inductive
type
[
vect
]
of
Coq
"vectors"
,
i
.
e
.
lists
of
a
given
size
.
*
)
(
**
**
Perfect
binary
trees
*
)
(
**
With
the
same
approach
,
we
could
represent
binary
tree
of
depth
[
n
]
which
are
perfect
(
e
.
g
.
all
leaves
are
at
the
same
depth
).
Here
we
put
a
[
nat
]
data
at
leaves
.
*
)
Fixpoint
bintree
(
A
:
Type
)
n
:=
match
n
with

0
=>
A

S
n
=>
(
bintree
A
n
*
bintree
A
n
)
%
type
end
.
Check
((
1
,
2
),(
3
,
4
))
:
bintree
nat
2.
(
**
Just
visualize
the
","
as
indicating
a
binary
node
.
*
)
(
**
For
instance
,
let
'
s
sum
all
data
in
such
a
bintree
*
)
Fixpoint
sumtree
n
:
bintree
nat
n
>
nat
:=
match
n
with

0
=>
fun
a
=>
a

S
n
=>
fun
'
(
g
,
d
)
=>
sumtree
n
g
+
sumtree
n
d
end
.
(
**
Now
,
if
we
want
to
put
some
data
on
the
nodes
rather
than
on
the
leaves
:
*
)
Set
Universe
Polymorphism
.
(
*
To
avoid
a
nasty
issue
with
universes
.
*
)
(
*
A
singleton
type
for
leaves
(
already
provided
in
Coq
)
:
Only
one
value
in
type
[
unit
],
namely
[
Tt
].
*
)
Inductive
unit
:
Type
:=
Tt
.
Fixpoint
bintree
'
(
A
:
Type
)
n
:=
match
n
with

0
=>
unit

S
n
=>
(
bintree
'
A
n
*
A
*
bintree
'
A
n
)
%
type
end
.
Check
((
Tt
,
1
,
Tt
),
2
,(
Tt
,
3
,
Tt
))
:
bintree
'
nat
2.
Fixpoint
sumtree
'
n
:
bintree
'
nat
n
>
nat
:=
match
n
with

0
=>
fun
Tt
=>
0

S
n
=>
fun
'
(
g
,
a
,
d
)
=>
sumtree
'
n
g
+
a
+
sumtree
'
n
d
end
.
(
**
Nota
:
here
we
used
Coq
triples
,
which
are
not
primitives
(
pairs
of
paires
).
We
could
also
have
defined
an
ad

hoc
inductive
type
for
triples
.
*
)
(
**
Once
again
,
using
a
[
Fixpoint
]
here
for
[
bintree
]
and
[
bintree
'
]
is
only
one
of
the
possible
solutions
,
we
could
also
have
defined
an
inductive
dependent
type
,
see
later
.
*
)
(
**
**
Functions
of
arity
[
n
]
*
)
(
**
Definition
of
a
type
of
functions
with
[
n
]
arguments
(
in
[
nat
])
and
a
answer
(
in
[
nat
]).
*
)
Fixpoint
narity
n
:=
match
n
with

0
=>
nat

S
n
=>
nat
>
narity
n
end
.
Compute
narity
5.
(
**
Example
of
usage
:
let
'
s
create
a
n

ary
addition
function
.
Beware
:
the
first
argument
is
[
n
],
indicating
how
many
more
arguments
are
to
be
expected
(
and
then
summed
).
But
this
first
argument
[
n
]
isn
'
t
added
itself
to
the
sum
.
*
)
(
**
In
a
first
time
,
it
is
convenient
to
consider
having
at
least
a
number
to
sum
,
this
way
we
can
use
it
as
an
accumulator
.
*
)
Fixpoint
narity_S_sum
n
:
narity
(
S
n
)
:=
match
n
with

0
=>
fun
a
=>
a

S
n
=>
fun
a
b
=>
narity_S_sum
n
(
a
+
b
)
end
.
(
**
We
can
now
generalize
for
any
possible
[
n
].
*
)
Definition
narity_sum
n
:
narity
n
:=
match
n
with

0
=>
0

S
n
=>
narity_S_sum
n
end
.
Compute
narity_sum
4
5
1
2
3
:
nat
.
(
*
4
numbers
to
sum
,
and
then
5
+
1
+
2
+
3
=
11
*
)
(
**
**
A
type
for
"number or Boolean"
*
)
(
**
We
can
control
(
for
instance
via
a
[
bool
])
whether
a
type
is
[
nat
]
or
[
bool
].
*
)
Definition
nat_or_bool
(
b
:
bool
)
:
Type
:=
if
b
then
nat
else
bool
.
(
**
An
example
of
value
in
this
type
.
*
)
Definition
value_nat_or_bool
(
b
:
bool
)
:
nat_or_bool
b
:=
match
b
return
(
nat_or_bool
b
)
with

true
=>
0

false
=>
false
end
.
(
**
This
[
match
b
...]
is
roughly
equivalent
to
[
if
b
then
0
else
false
]
but
writing
a
[
if
]
here
would
not
give
Coq
enough
details
.
Even
putting
the
explicit
output
type
isn
'
t
enough
,
we
add
to
use
an
extra
syntax
[
return
...]
to
help
Coq
.
*
)
(
**
This
[
nat_or_bool
]
type
is
not
so
helpful
in
itself
.
But
we
could
use
a
similar
idea
when
writing
the
interpretor
of
an
abstract
language
,
where
the
interpretation
results
may
be
in
several
types
.
Fixpoint
expected_type
(
e
:
abstract_expression
)
:
Type
:=
...
Fixpoint
interpret
(
e
:
abstract_expression
)
:
expected_type
e
:=
...
*
)
(
**
To
compare
,
a
more
basic
way
to
proceed
(
non

dependently
)
would
be
an
inductive
type
representing
all
the
possible
outcomes
,
see
below
.
But
accessing
values
in
this
type
is
cumbersome
(
just
like
accessing
the
value
in
an
[
option
]
type
).
*
)
Inductive
natOrbool
:=

Nat
:
nat
>
natOrbool

Bool
:
bool
>
natOrbool
.
(
*
What
if
we
have
a
[
natOrBool
]
and
want
to
access
the
[
nat
]
in
it
?
What
if
we
are
actually
in
front
of
a
[
bool
]
?
That
would
be
an
error
we
would
have
to
handle
,
while
the
solution
based
on
dependent
types
is
immune
from
this
issue
.
*
)
(
**
**
Inductive
dependent
types
*
)
(
**
Vectors
*
)
(
**
This
is
another
encoding
of
n

uples
.
We
almost
reuse
the
inductive
definition
of
lists
,
but
we
add
an
extra
parameter
representing
the
length
of
the
list
.
Cf
.
the
Coq
standard
library
,
file
[
Vector
.
v
]
*
)
Inductive
vect
(
A
:
Type
)
:
nat
>
Type
:=

Vnil
:
vect
A
0

Vcons
n
:
A
>
vect
A
n
>
vect
A
(
S
n
).
Arguments
Vnil
{
A
}
.
(
*
Vnil
and
Vcons
with
implicit
domains
*
)
Arguments
Vcons
{
A
}
.
(
**
Codage
du
triplet
(
0
,
2
,
3
)
:
*
)
Check
(
Vcons
2
0
(
Vcons
1
2
(
Vcons
0
3
Vnil
))).
(
**
The
other
numbers
2
0
1
indicates
the
lengths
of
each
sub

vector
.
Since
this
is
pretty
predictable
,
and
hence
"boring"
,
we
could
even
hide
the
[
n
]
argument
of
[
Vcons
].
*
)
Arguments
Vcons
{
A
}
{
n
}
.
Check
(
Vcons
0
(
Vcons
2
(
Vcons
3
Vnil
))).
(
**
But
this
[
n
]
argument
is
still
there
internally
.
*
)
Set
Printing
All
.
Check
(
Vcons
0
(
Vcons
2
(
Vcons
3
Vnil
))).
Check
(
cons
0
(
cons
2
(
cons
3
nil
))).
Unset
Printing
All
.
(
**
Conversion
bewteen
vectors
and
regular
lists
.
*
)
Require
Import
List
.
Import
ListNotations
.
Fixpoint
v2l
{
A
}
{
n
}
(
v
:
vect
A
n
)
:
list
A
:=
match
v
with

Vnil
=>
[]

Vcons
x
v
=>
x
::
(
v2l
v
)
end
.
Fixpoint
l2v
{
A
}
(
l
:
list
A
)
:
vect
A
(
length
l
)
:=
match
l
with

[]
=>
Vnil

x
::
l
=>
Vcons
x
(
l2v
l
)
end
.
(
**
The
following
definition
(
re

)
computing
the
length
of
a
vector
is
actually
useless
since
we
already
known
this
length
,
it
is
the
parameter
[
n
]
mentionned
in
the
type
[
vect
].
Later
,
we
could
prove
that
[
forall
A
n
(
v
:
vect
A
n
),
length
v
=
n
].
*
)
Fixpoint
length
{
A
}
{
n
}
(
v
:
vect
A
n
)
:
nat
:=
match
v
with

Vnil
=>
0

Vcons
_
v
=>
1
+
length
v
end
.
(
**
With
such
vectors
,
we
avoid
the
usual
issues
encountered
when
defining
[
head
]
and
alii
on
listes
.
Indeed
,
we
can
specify
that
we
want
to
operate
only
on
non

empty
vectors
,
since
they
have
a
[
Vect
A
(
S
n
)]
type
for
some
[
n
].
*
)
(
**
On
lists
:
*
)
Definition
head
{
A
}
(
l
:
list
A
)
(
default
:
A
)
:
A
:=
match
l
with

[]
=>
default

x
::
_
=>
x
end
.
(
**
Or
:
*
)
Definition
head_opt
{
A
}
(
l
:
list
A
)
:
option
A
:=
match
l
with

[]
=>
None

x
::
_
=>
Some
x
end
.
(
**
On
vectors
:
*
)
Definition
Vhead
{
A
}
{
n
}
(
v
:
vect
A
(
S
n
))
:
A
:=
match
v
with

Vcons
x
_
=>
x
end
.
(
**
The
match
branch
for
[
Vnil
]
just
vanished
,
and
Coq
didn
'
t
complained
about
a
missing
case
!
Actually
,
there
does
exist
such
a
case
internally
,
filled
by
Coq
with
a
somewhat
arbitrary
code
,
that
will
never
be
accessed
during
computations
.
*
)
Print
Vhead
.
Compute
Vhead
((
Vcons
0
(
Vcons
2
(
Vcons
3
Vnil
)))).
Fail
Compute
Vhead
Vnil
.
(
**
Concatenating
vectors
Everything
goes
smoothly
here
,
since
this
function
mimics
the
equations
of
the
addition
on
[
nat
].
*
)
Print
Nat
.
add
.
(
*
0
+
m
=
m
S
p
+
m
=
S
(
p
+
m
)
*
)
Fixpoint
Vapp
{
A
}
{
n
m
}
(
v
:
vect
A
n
)
(
v
'
:
vect
A
m
)
:
vect
A
(
n
+
m
)
:=
match
v
with

Vnil
=>
v
'
(
*
of
type
vect
A
m
=
vect
A
(
0
+
m
)
=
vect
A
(
n
+
m
)
here
*
)

Vcons
x
v
=>
Vcons
x
(
Vapp
v
v
'
)
(
*
of
type
vect
A
(
S
(
p
+
m
))
=
vect
A
((
S
p
)
+
m
)
=
vect
A
(
n
+
m
)
here
*
)
end
.
(
**
Alas
,
for
other
algorithms
,
we
arent
'
so
lucky
.
For
instance
,
consider
the
(
naive
)
definition
of
the
reverse
of
a
vector
.
A
direct
attempt
is
rejected
as
badly

typed
.
Coq
would
need
to
know
that
[
vect
A
(
n
+
1
)]
is
equal
to
[
vect
A
(
1
+
n
)].
This
is
provably
true
,
but
not
computationally
true
(
a
.
k
.
a
"convertible"
in
the
Coq
jargon
).
More
precisely
,
we
cannot
have
[
n
+
1
=
1
+
n
]
by
just
the
above
(
Peano
)
equations
,
this
would
later
need
a
proof
by
induction
.
*
)
Fail
Fixpoint
Vrev
{
A
}
{
n
}
(
v
:
vect
A
n
)
:
vect
A
n
:=
match
v
with

Vnil
=>
Vnil

Vcons
x
v
=>
Vapp
(
Vrev
v
)
(
Vcons
x
Vnil
)
end
.
(
**
A
first
solution
:
use
the
Coq
equality
to
"cast"
between
vectors
of
equal
lengths
.
We
'
ll
detail
more
later
how
Coq
represents
its
logical
equality
,
for
now
let
'
s
just
say
that
it
'
s
internally
a
syntactic
correspondence
between
the
two
side
of
the
equation
.
Hence
"matching"
over
this
proof
[
h
]
of
type
[
n
=
m
]
would
formally
"equate"
[
n
]
with
[
m
].
And
all
it
well
.
*
)
Print
eq
.
Definition
Vcast
{
A
}
{
n
}
{
m
}
(
v
:
vect
A
n
)(
h
:
n
=
m
)
:
vect
A
m
:=
match
h
with

eq_refl
=>
v
end
.
(
**
But
for
defining
our
[
Vrev
],
we
need
a
proof
of
[
n
+
1
=
1
+
n
].
*
)
Require
Import
Arith
.
SearchPattern
(
_
+
1
=
_
).
(
*
That
'
s
[
Nat
.
add_1_r
]
we
need
.
*
)
Fixpoint
Vrev
{
A
}
{
n
}
(
v
:
vect
A
n
)
:
vect
A
n
:=
match
v
with

Vnil
=>
Vnil

Vcons
x
v
=>
Vcast
(
Vapp
(
Vrev
v
)
(
Vcons
x
Vnil
))
(
Nat
.
add_1_r
_
)
end
.
(
**
Issue
:
all
computations
with
this
[
Vrev
]
is
blocked
,
since
they
cannot
access
Coq
standard
proof
of
[
Nat
.
add_1_r
]
(
such
a
proof
is
said
to
be
"opaque"
).
*
)
Compute
Vrev
(
Vcons
1
(
Vcons
2
(
Vcons
3
Vnil
))).
(
**
Solution
inside
the
solution
:
do
the
proof
ourself
,
in
a
"transparent"
way
.
Forget
about
this
proof
for
the
moment
.
*
)
Lemma
add_1_r
n
:
n
+
1
=
S
n
.
Proof
.
induction
n
;
simpl
;
trivial
.
now
f_equal
.
Defined
.
(
**
And
not
the
usual
Qed
keyword
,
that
would
be
opaque
!
*
)
Fixpoint
Vrev_transp
{
A
}
{
n
}
(
v
:
vect
A
n
)
:
vect
A
n
:=
match
v
with

Vnil
=>
Vnil

Vcons
x
v
=>
Vcast
(
Vapp
(
Vrev_transp
v
)
(
Vcons
x
Vnil
))
(
add_1_r
_
)
end
.
Compute
Vrev_transp
(
Vcons
1
(
Vcons
2
(
Vcons
3
Vnil
))).
(
**
Another
solution
:
another
definition
of
[
Vcast
],
way
more
complex
,
that
proceed
by
induction
over
the
vector
instead
of
doing
an
induction
over
the
equality
proof
.
You
can
skip
the
(
ugly
)
details
of
this
function
[
Vcast2
],
just
notice
that
it
does
exist
.
*
)
Definition
Vcast2
:
forall
{
A
m
}
(
v
:
vect
A
m
)
{
n
}
,
m
=
n
>
vect
A
n
.
Proof
.
refine
(
fix
cast
{
A
m
}
(
v
:
vect
A
m
)
{
struct
v
}
:=
match
v
in
vect
_
m
'
return
forall
n
,
m
'
=
n
>
vect
A
n
with

Vnil
=>
fun
n
=>
match
n
with

0
=>
fun
H
=>
Vnil

S
_
=>
fun
H
=>
False_rect
_
_
end

Vcons
h
w
=>
fun
n
=>
match
n
with

0
=>
fun
H
=>
False_rect
_
_

S
n
'
=>
fun
H
=>
Vcons
h
(
cast
w
n
'
(
f_equal
pred
H
))
end
end
);
discriminate
.
Defined
.
(
**
Computing
with
this
[
Vcast2
]
means
deconstructing
the
whole
vector
[
v
]
before
reconstructing
everything
(
but
with
a
length
expressed
with
the
other
side
of
the
equality
).
*
)
Print
Vcast2
.
(
*
fix
...
match
v
...
Vil
=>
...
Vnil
...

Vcons
n
h
w
=>
...
Vcons
h
...
*
)
(
**
To
get
the
underlying
algorithm
behind
all
these
details
,
a
nice
way
it
to
use
the
"extraction"
tool
,
that
convert
a
Coq
definition
into
some
corresponding
OCaml
code
,
and
drop
on
the
fly
all
logical
parts
(
such
as
the
equalities
).
More
on
that
to
come
later
.
*
)
Require
Extraction
.
Extraction
Vcast2
.
(
**
And
finally
:
*
)
Fixpoint
Vrev2
{
A
}
{
n
}
(
v
:
vect
A
n
)
:
vect
A
n
:=
match
v
with

Vnil
=>
Vnil

Vcons
x
v
=>
Vcast2
(
Vapp
(
Vrev2
v
)
(
Vcons
x
Vnil
))
(
Nat
.
add_1_r
_
)
end
.
(
**
And
we
can
compute
here
,
even
with
Coq
standard
proof
of
[
n
+
1
=
1
+
n
].
*
)
Compute
Vrev2
(
Vcons
1
(
Vcons
2
(
Vcons
3
Vnil
))).
(
**
As
a
conclusion
,
programming
with
dependent
type
may
help
a
lot
by
ruling
out
some
impossible
cases
and
producing
functions
with
rich
types
describing
precisely
the
current
situation
.
But
it
is
not
always
convenient
to
proceed
in
this
style
,
and
may
lead
you
to
some
definitional
nightmare
such
as
this
[
Vrev
].
*
)
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