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cpdt
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e78f5c70
Commit
e78f5c70
authored
Apr 28, 2010
by
Adam Chlipala
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Typo fix in Equality, found proof-reading tutorial
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src/Equality.v
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e78f5c70
...
@@ -669,7 +669,7 @@ End fhapp'.
...
@@ -669,7 +669,7 @@ End fhapp'.
(
*
EX
:
Show
that
the
approaches
based
on
K
and
JMeq
are
equivalent
logically
.
*
)
(
*
EX
:
Show
that
the
approaches
based
on
K
and
JMeq
are
equivalent
logically
.
*
)
(
*
begin
thide
*
)
(
*
begin
thide
*
)
(
**
Assuming
axioms
(
like
axiom
K
and
[
JMeq_eq
])
is
a
hazardous
business
.
The
due
diligence
associated
with
it
is
necessarily
global
in
scope
,
since
two
axioms
may
be
consistent
alone
but
inconsistent
together
.
It
turns
out
that
all
of
the
major
axioms
proposed
for
reasoning
about
equality
in
Coq
are
logically
equivalent
,
so
that
we
only
need
to
pick
one
to
assert
without
proof
.
In
this
section
,
we
demonstrate
this
by
showing
how
each
the
previous
two
sections
'
approaches
reduces
to
the
other
logically
.
(
**
Assuming
axioms
(
like
axiom
K
and
[
JMeq_eq
])
is
a
hazardous
business
.
The
due
diligence
associated
with
it
is
necessarily
global
in
scope
,
since
two
axioms
may
be
consistent
alone
but
inconsistent
together
.
It
turns
out
that
all
of
the
major
axioms
proposed
for
reasoning
about
equality
in
Coq
are
logically
equivalent
,
so
that
we
only
need
to
pick
one
to
assert
without
proof
.
In
this
section
,
we
demonstrate
this
by
showing
how
each
of
the
previous
two
sections
'
approaches
reduces
to
the
other
logically
.
To
show
that
[
JMeq
]
and
its
axiom
let
us
prove
[
UIP_refl
]
,
we
start
from
the
lemma
[
UIP_refl
'
]
from
the
previous
section
.
The
rest
of
the
proof
is
trivial
.
*
)
To
show
that
[
JMeq
]
and
its
axiom
let
us
prove
[
UIP_refl
]
,
we
start
from
the
lemma
[
UIP_refl
'
]
from
the
previous
section
.
The
rest
of
the
proof
is
trivial
.
*
)
...
...
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