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Proof step 33 from proof multi-at-other (previous step)

(addmultiarrays (firstn(# refs - 1 + i→n(countrefs(ints1)) * i→n(i1), 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 = 2 + i→n(countrefs(ints1)) * i→n(i1) + i→n(i2) * i→n(countrefs(ints1)) * i→n(i1) + i→n(i2), lastn(1 + i→n(countrefs(ints1)) * i→n(i1), refs) = r0 ' + refs1, 0 < i0, i2 < i0, 0 ≤ i2, 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])) ⊦ r ≠ r0, (addmultiarrays(r0 ' + refs1, i1, [], ty, i1 ' + ints1, st)[r -- sk]) = st[r -- sk], 0 = i0, (~1) = i2, ints1 = [], 0 = i1, r ∈ refs, 0 ∈ ints1 .butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 32:
⊦ ¬ r0 ∈ refs ∧ # refs = 2 + n * m + m0 → lastn(1 + n * m, refs) ≠ r0 ' + refs1

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m0 * m * n = (m0 * m) * n])]]