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Proof step 19 from proof multg-at-ref (previous step)

∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#([]) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ (addarraymultlist(r, ty, st, refs, [], ints0, i)[r0 -- sk]) = st[r0 -- sk], r = r0, r0 ∈ refs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 18:
⊦ # ints = 0 → addarraymultlist(r, ty, st, refs, ints, ints0, i) = st

Rule argument

[crewrite-arg: addarraymultlist_zero , F , [i, ints0, ints, refs, st, ty, r] → [i, ints0, [], refs, st, ty, r] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # [] = 0])], [EXTRATERMS : (i, ints0, [], refs, st, ty, r)]]