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Proof step 21 from proof multi-in-other (previous step)

# refs2 = 1 + # refs1 + i→n(i2) * # refs1 + i→n(i2), lastn(1 + # refs1, r1 ' + refs2) = r0 ' + refs1, r -- sk ∈ addmultiarrays (r1 ' + firstn(i→n(i2) * # refs1 + i→n(i2), refs2), 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), 1 + # refs1 = i→n(countrefs(i1 ' + ints1)), # refs1 ≤ # refs2, posints(ints1), 0 ≤ i1, 0 < i0, i2 < i0, 0 ≤ i2, ∀ st, ty, acc, sk, ints, i, r, refs0. # refs0 < # (r1 + refs2) → ( true → ( ¬ r ∈ refs0 ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → (r -- sk ∈ addmultiarrays(refs0, i, acc, ty, ints, st) ↔ r -- sk ∈ st))) ⊦ r ≠ r0, r -- sk ∈ addmultiarrays(r0 ' + refs1, i1, [], ty, i1 ' + ints1, st) ↔ r -- sk ∈ st, 0 = i0, (~1) = i2, r = r1, 0 ∈ (i1 ' + ints1) .butlast, r ∈ refs2

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])]]