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Proof step 72 from proof multg-at-ref (previous step)

i→n(countrefs2(ints)) ≤ # refs, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(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, ints, 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 1:
⊦ # ints = 1 ∧ ints .first ≤ 0 → addarraymultlist(r, ty, st, refs, ints, ints0, i) = (st[r -- _type, typeval(mktype_from_dims(ty, n→i(# ints) + i))][r -- _length, intval(ints0 .first)])
Proof step 71:
¬ (# ints = 1 ∧ ints .first ≤ 0), i→n(countrefs2(ints)) ≤ # refs, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(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, ints, ints0, i)[r0 -- sk]) = st[r0 -- sk], r = r0, r0 ∈ refs

Rule argument

[crewrite-arg: addarraymultlist_one_0 , F , [i, ints0, ints, refs, st, ty, r] → [i, ints0, ints, refs, st, ty, r] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ≠ r0 → (st[r0 -- sk0, val][r -- sk]) = st[r -- sk]])], [EXTRATERMS : (i, ints0, ints, refs, st, ty, r)]]