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Proof step 59 from proof multg-at-ref (previous step)

refs = r1 + refs0 ∧ refs ≠ [], 2 ≤ i0, i→n(i0 * (countrefs2(ints2) + 1)) ≤ # refs, 0 ≤ i0, 0 < i0, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(i0 + ints2) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ r = r0, ints2 = [], (addarraymultlist (r1, ty, addarraymultlist (r, ty, st[r -- (i0 - 1) ', refval(r1)][r -- _type, typeval(mktype_from_dims(ty, (1 + n→i(# ints2)) + i))][r -- _length, intval(ints0 .first)], refs0, (i0 - 1) ' + ints2, ints0, i), restn(i→n(countrefs2((i0 - 1) ' + ints2)), refs0), ints2, ints0 .rest, i) [r0 -- sk ] ) = st[r0 -- sk], r0 ∈ refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 58:
0 ≤ i0, 0 < i0, 2 ≤ i0, i→n(i0 * (countrefs2(ints2) + 1)) ≤ 1 + # refs0, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(i0 + ints2) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ r = r0, ints2 = [], (addarraymultlist (r1, ty, addarraymultlist (r, ty, st[r -- (i0 - 1) ', refval(r1)][r -- _type, typeval(mktype_from_dims(ty, (1 + n→i(# ints2)) + i))][r -- _length, intval(ints0 .first)], refs0, (i0 - 1) ' + ints2, ints0, i), restn(i→n(countrefs2((i0 - 1) ' + ints2)), refs0), ints2, ints0 .rest, i) [r0 -- sk ] ) = st[r0 -- sk], r0 = r1, r0 ∈ refs0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ r0 ' + refs ↔ r ≠ r0 ∧ ¬ r ∈ refs], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]