Commit 4d5282fa by Björn Brandenburg

### protect TODO comments from spell checker

parent 165a18f2
Pipeline #69077 passed with stage
in 18 minutes and 16 seconds
 ... ... @@ -35,8 +35,8 @@ Section SumsOverSequences. (** We start showing that having every member of [r] equal to zero is equivalent to having the sum of all the elements of [r] equal to zero, and vice-versa. *) (* TODO: PR MathComp this should probably be named sum_nat_eq0, but there is already a sum_nat_eq0 that is less generic? *) this should probably be named [sum_nat_eq0], but there is already a [sum_nat_eq0] that is less generic? *) Lemma sum_nat_eq0_nat : (\sum_(i <- r | P i) F i == 0) = all (fun x => F x == 0) [seq x <- r | P x]. Proof. ... ... @@ -90,7 +90,7 @@ Section SumsOverSequences. guarantee that the set inclusion [r <= rs] implies the actually required multiset inclusion. *) (* TODO: PR MathComp - add a condition P i *) - add a condition [P i] *) Lemma leq_sum_sub_uniq (rs : seq I) : uniq r -> {subset r <= rs} -> \sum_(i <- r) F i <= \sum_(i <- rs) F i. ... ... @@ -132,7 +132,7 @@ Section SumsOverSequences. (** In the same way, if [E1] equals [E2] in all the points considered above, then also the two sums will be identical. *) (* TODO: PR MathComp - generalize as eq_big_seq_cond (nothing specific to addn here) - generalize as [eq_big_seq_cond] (nothing specific to [addn] here) - replace == with = ? *) Lemma eq_sum_seq: (forall i, i \in r -> P i -> E1 i == E2 i) -> ... ... @@ -147,7 +147,7 @@ Section SumsOverSequences. points, then the sum of [E] conditioned by [P2] will dominate the sum of [E] conditioned by [P1]. *) (* TODO: PR MathComp - maybe leq_sum_seq above should be leq_sum_seqr and this one leq_sum_seql *) - maybe [leq_sum_seq] above should be [leq_sum_seqr] and this one [leq_sum_seql] *) Lemma leq_sum_seq_pred: (forall i, i \in r -> P1 i -> P2 i) -> \sum_(i <- r | P1 i) E i <= \sum_(i <- r | P2 i) E i. ... ...
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