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Explanation about the axiom (substitution lemma)
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carlosolarte committed Jun 25, 2017
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Expand Up @@ -199,27 +199,26 @@ Module SqSystems (DT : Eqset_dec_pol).
Qed.

(**
Since there are no free variables in our encoding, we cannot
prove directly the usual substitution lemma: if there is a proof
with a fresh variable x, then there is a proof, of the same
height, for a given term t. This axiom (and the similar ones for
the other systems) are introduced to cope with the following
cases:
Since there are no free variables in our encoding, we cannot prove
directly the usual substitution lemma: if there is a proof with a
fresh variable x, then there is a proof, of the same height for any
term t. The following axiom (and the similar ones for the other
systems) are introduced to cope with proofs of the form:
<<
H: forall x:Term, exists n:nat, |- Gamma, Subst FX x
----------------------------------------------------
exists n:nat, |- Gamma, F{ FX}
G: exists n:nat, |- Gamma, F{ FX}
>>
The hypothesis [H] is the result of our encoding of the LL universal
quantifier as the (dependent type) constructor [forall] in Coq. In the
usual sequent calculi presentation, that hypothesis is stronger:
#<i>for any fresh variable x, there is a proof of height n</i>#. Then,
it suffices to generalize H with a fresh variable, and then, from
alpha conversion, we can use exactly the same fresh variable to
conclude the goal.
The hypothesis [H] results in inductive proofs where the principal
formula is (the LL universal quantifier} [F{ FX}]. Note that we cannot
conclude the goal [G] from [H] since our hypothesis is weaker than the
similar one in pencil/paper proofs. More precisely, in a paper proof,
we can generalize [H] with a fresh variable [x]. Then, there exists
[n] s.t. [n |- Gamma, Subst Fx x]. By using the substitution lemma,
for any [y], it must be the case [n |- Gamma, Subst Fx y] and we can
easily conclude the goal [G].
*)
Axiom ax_subs_prop: forall B L M FX (P:nat -> Prop), (forall x : Term, exists n : nat, (P n) /\ n |-F- B; L; UP (Subst FX x :: M)) -> exists n, (P n) /\ forall x, n |-F- B; L; UP (Subst FX x :: M) .

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