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Proof step 26 from proof multi-length-first (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
⊦ countrefs(ints) ≠ 0
Proof step 25:
lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs) = r0 + refs1 ∧ lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs) ≠ [], # refs = i→n(countrefs(i0 ' + ints1)) * i→n(i1) + i→n(countrefs(i0 ' + ints1)), posints(ints1), 0 ≤ i0, 0 ≤ n→i(# acc) + i1, 0 ≤ i1, ∀ 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))) ⊦ (addmultiarrays (firstn(1 + # refs - i→n(countrefs(i0 ' + ints1)), r ' + refs), i1, r0 ' + acc, ty, (n→i(# acc) + i1 + 1) ' + i0 ' + ints1, addmultiarrays(lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs), i0, [], ty, i0 ' + ints1, st)) [r -- _length ] ) = intval(n→i(# acc) + i1 + 1), (~1) = n→i(# acc) + i1, (~1) = i1, 0 ∈ (i0 ' + ints1) .butlast

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs), r0, refs1] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ lastn(n, refs) = [] ↔ n = 0], [CSIMPSEQ : ⊦ i→n(i) = 0 ↔ i = 0], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (lastn(i→n(countrefs(i0 ' + ints1)), r ' + refs), r0, refs1)]]