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Proof step 50 from proof multi-at-other (previous step)

ints0 = i1 + ints1, # refs = i→n(countrefs(i ' + ints0)), posints(ints0), 0 < i0, i ≤ i0, 0 < i, ∀ 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])) ⊦ ints0 = [], 0 = i0, i = 0, (addmultiarrays (firstn(# refs - i→n(countrefs(ints0)), refs), i - 1, lastn(i→n(countrefs(ints0)), refs) .first ' + acc, ty, i0 ' + ints0, addmultiarrays(lastn(i→n(countrefs(ints0)), refs), i1, [], ty, ints0, st)) [r -- sk ] ) = st[r -- sk], r ∈ refs, 0 ∈ ints0 .butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 49:
# refs = i→n(countrefs(i ' + i1 ' + ints1)), 0 < i0, i ≤ i0, 0 < i, posints(ints1), 0 ≤ i1, ∀ 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])) ⊦ 0 = i0, i = 0, (addmultiarrays (firstn(# refs - i→n(countrefs(i1 ' + ints1)), refs), i - 1, lastn(i→n(countrefs(i1 ' + ints1)), refs) .first ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(lastn(i→n(countrefs(i1 ' + ints1)), refs), i1, [], ty, i1 ' + ints1, st)) [r -- sk ] ) = st[r -- sk], r ∈ refs, 0 ∈ (i1 ' + ints1) .butlast

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ posints(i ' + ints) ↔ 0 ≤ i ∧ posints(ints)], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ i ' + ints ≠ []], [CSIMPSEQ : ⊦ i + ints = i ' + ints])]]