(MEMBER (FOO A) (APPEND (BAR B) (MUM C)))
Major Section: INTRODUCTION-TO-THE-THEOREM-PROVER
What rules come to mind when looking at the following subterm of a Key Checkpoint? Think of strong rules (see strong-rewrite-rules).
(MEMBER (FOO A) (APPEND (BAR B) (MUM C)))
Since (append x y) contains all the members of x and all the members
of y, e is a member of (append x y) precisely when e is a
member of x or of y. So a strong statement of this is:
(defthm member-append-strong-false
(equal (member e (append x y))
(or (member e x)
(member e y))))
However, this is not a theorem because member is not Boolean.
(Member e x), for example, returns the first tail of x that starts
with e, or else nil. To see an example of this formula that
evaluates to nil, let
e = 3 x = '(1 2 3) y = '(4 5 6).Then the left-hand side,
(member e (append x y)) evaluates to (3 4 5 6) while
the right-hand side evaluates to (3).However, the two sides are propositionally equivalent (both either nil
or non-nil together). So this is a useful :rewrite
rule:
(defthm member-append-strong
(iff (member e (append x y))
(or (member e x)
(member e y)))).
It tells the system that whenever it encounters an instance of
(MEMBER e (APPEND x y)) in a propositional occurrence (where only
its truthvalue is relevant), it should be replaced by this
disjunction of (MEMBER e x) and (MEMBER e y).The following two formulas are true but provide much weaker rules and we would not add them:
(implies (member e x) (member e (append x y))) (implies (member e y) (member e (append x y)))because they each cause the system to backchain upon seeing
(MEMBER e (APPEND x y))
expressions and will not apply unless one of the two side-conditions can be established.There is a rewrite rule that is even stronger than member-append-strong.
It is suggested by the counterexample, above, for the EQUAL version of the rule.
(defthm member-append-really-strong
(equal (member e (append x y))
(if (member e x)
(append (member e x) y)
(member e y))))
While member-append-strong only rewrites member-append expressions
occurring propositionally, the -really-strong version rewrites every
occurrence.However, this rule will be more useful than member-append-strong only
if you have occurrences of member in non-propositional places. For example,
suppose you encountered a term like:
(CONS (MEMBER e (APPEND x y)) z).Then the
-strong rule does not apply but the -really-strong rule does.Furthermore, the -really-strong rule, by itself, is not quite as good as
the -strong rule in propositional settings! For example, if you have proved
the -really-strong rule, you'll notice that the system still has to use
induction to prove
(IMPLIES (MEMBER E A)
(MEMBER E (APPEND B A))).
The -really-strong rule would rewrite it to
(IMPLIES (MEMBER E A)
(IF (MEMBER E A)
(APPEND (MEMBER E A) B)
(MEMBER E B)))
which would further simplify to
(IMPLIES (MEMBER E A)
(APPEND (MEMBER E A) B))
What lemma does this suggest? The answer is the rather odd:
(implies x (append x y))which rewrites propositional occurrences of
(APPEND x y) to T if
x is non-nil. This is an inductive fact about append.A problem with the -really-strong rule is that it transforms even
propositional occurrences of member into mixed propositional and
non-propositional occurrences.
(defthm member-append-really-strong
(equal (member e (append x y)) ; <-- even if this is a propositional occurrence
(if (member e x)
(append (member e x) y) ; <-- the member in here is not!
(member e y))))
So if you are using the -really-strong lemma in a situation in which
all your member expressions are used propositionally, you'll suddenly
find yourself confronted with non-propositional uses of member.Our advice is not to use the -really-strong version unless your application is
inherently using member in a non-propositional way.
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