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Proof step 28 from proof multi-length-first (previous step)

# refs + 1 = i→n(countrefs(i ' + i0 ' + ints1)), 0 < n→i(# acc) + i, 0 < i, posints(ints1), 0 ≤ i0, ∀ st, ty, r, acc, ints, i, refs0. # refs0 < # refs → ( true → ( # refs0 + 1 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first ∧ n→i(# acc) + i = ints .first → (addmultiarrays(r ' + refs0, i, acc, ty, ints, st)[r -- _length]) = intval(ints .first))) ⊦ 0 = n→i(# acc) + i, i = 0, (addmultiarrays (firstn(1 + # refs - i→n(countrefs(i0 ' + ints1)), r ' + refs), i - 1, lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs) .first ' + acc, ty, (n→i(# acc) + i) ' + i0 ' + ints1, addmultiarrays(lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs), i0, [], ty, i0 ' + ints1, st)) [r -- _length ] ) = intval(n→i(# acc) + i), 0 ∈ (i0 ' + ints1) .butlast

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i, 1, i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i, 1, i1)]]