diff -r f424e164d752 -r 5e315defb038 src/HOL/Analysis/Infinite_Products.thy --- a/src/HOL/Analysis/Infinite_Products.thy Sat May 09 17:20:04 2020 +0000 +++ b/src/HOL/Analysis/Infinite_Products.thy Mon May 11 11:15:41 2020 +0100 @@ -1185,7 +1185,7 @@ lemma raw_has_prod_Suc_iff: "raw_has_prod f M (a * f M) \ raw_has_prod (\n. f (Suc n)) M a \ f M \ 0" proof - have "raw_has_prod f M (a * f M) \ (\i. \j\Suc i. f (j+M)) \ a * f M \ a * f M \ 0" - by (subst LIMSEQ_Suc_iff) (simp add: raw_has_prod_def) + by (subst filterlim_sequentially_Suc) (simp add: raw_has_prod_def) also have "\ \ (\i. (\j\i. f (Suc j + M)) * f M) \ a * f M \ a * f M \ 0" by (simp add: ac_simps atMost_Suc_eq_insert_0 image_Suc_atMost prod.atLeast1_atMost_eq lessThan_Suc_atMost del: prod.cl_ivl_Suc) @@ -1238,7 +1238,7 @@ proof - have "(\n. \i\{0..Suc n}. f (i + M)) \ L" using M_L - apply (subst (asm) LIMSEQ_Suc_iff[symmetric]) + apply (subst (asm) filterlim_sequentially_Suc[symmetric]) using atLeast0AtMost by auto then have "(\n. f M * (\i\{0..n}. f (Suc (i + M)))) \ L" apply (subst (asm) prod.atLeast0_atMost_Suc_shift)