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Proof step 27 from proof multi-at-other (previous step)

# refs = n + i→n(countrefs(i1 ' + ints1)) ∧ i→n(countrefs(i1 ' + ints1)) ≤ # refs, (addmultiarrays(firstn(n, refs), i2, r0 ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(r0 ' + refs1, i1, [], ty, i1 ' + ints1, st))[r -- sk]) = (addmultiarrays(r0 ' + refs1, i1, [], ty, i1 ' + ints1, st)[r -- sk]), # refs = 1 + i→n(countrefs(i1 ' + ints1)) * i→n(i2) + i→n(countrefs(i1 ' + ints1)), lastn(i→n(countrefs(i1 ' + ints1)), refs) = r0 ' + refs1, posints(ints1), 0 ≤ i1, 0 < i0, i2 < i0, 0 ≤ i2, ∀ sk, st, ty, acc, ints, i, r, refs0. # refs0 < # refs → ( true → ( ¬ r ∈ refs0 ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → (addmultiarrays(refs0, i, acc, ty, ints, st)[r -- sk]) = st[r -- sk])) ⊦ (addmultiarrays(r0 ' + refs1, i1, [], ty, i1 ' + ints1, st)[r -- sk]) = st[r -- sk], 0 = i0, (~1) = i2, r ∈ refs, 0 ∈ (i1 ' + ints1) .butlast, r ≠ r0 ∧ ¬ r ∈ refs1 ∧ 1 + # refs1 = i→n(countrefs(i1 ' + ints1)), ints1 ≠ [] ∧ 0 ≠ i1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ # refs = 0 ↔ refs = []], [CSIMPSEQ : ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n], [CSIMPSEQ : ⊦ m0 * 1 = m0], [CSIMPSEQ : ⊦ r ' = r0 ' + refs ↔ r = r0 ∧ refs = []], [CSIMPSEQ : ⊦ lastn(1, refs) = refs .last '], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ ints + [] = ints], [CSIMPSEQ : ⊦ n ≤ # refs ∧ r ∈ refs0 ∧ ¬ r ∈ refs → lastn(n, refs) ≠ r0 ' + refs0], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ n ≤ # refs ∧ ¬ r ∈ refs → lastn(n, refs) ≠ r ' + refs0])]]