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Proof step 15 from proof multi-at-other (previous step)

# 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])) ⊦ ¬ r ∈ firstn(# refs - i→n(countrefs(i1 ' + ints1)), refs) ∧ # refs - i→n(countrefs(i1 ' + ints1)) = i→n(countrefs(i1 ' + ints1)) * i→n(i2) + 1, (addmultiarrays (firstn(# refs - i→n(countrefs(i1 ' + ints1)), refs), i2, r0 ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(r0 ' + refs1, i1, [], ty, i1 ' + ints1, st)) [r -- sk ] ) = st[r -- sk], 0 = i0, (~1) = i2, r ∈ refs, 0 ∈ (i1 ' + ints1) .butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]