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Proof step 29 from proof multi-at-other (previous step)

¬ (# (i1 ' + ints1) > 1 ∧ (i1 ' + ints1) .first > 0), (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 ] ) = (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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
(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 ] ) = (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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ≤ j → (j ≤ i ↔ i = j)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ i > j ↔ j < i], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ 1 = m + 1 ↔ m = 0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)], [CSIMPSEQ : ⊦ m > n ↔ n < m], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0])]]