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Proof step 10 from proof multg-at-ref (previous step)

refs = r1 + refs0, i→n(countrefs2(ints2)) < # refs, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(1 + 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, refs = [], ints2 = [], (addarraymultlist (r1, ty, st[r -- 0 ', refval(r1)][r -- _type, typeval(mktype_from_dims(ty, (1 + n→i(# ints2)) + i))][r -- _length, intval(ints0 .first)], 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 9:
i→n(countrefs2(ints2)) ≤ # refs0, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(1 + 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, st[r -- 0 ', refval(r1)][r -- _type, typeval(mktype_from_dims(ty, (1 + n→i(# ints2)) + i))][r -- _length, intval(ints0 .first)], 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 : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ # r ' = 0 +1], [CSIMPSEQ : ⊦ # (refs + refs0) = # refs + # refs0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ r ' + refs ≠ []], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]