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Proof step 19 from proof multi-refin (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 18:
refs = r0 + refs2 ∧ refs ≠ [], # refs = (i→n(countrefs(i1 ' + ints1)) * i→n(i2) + 1) + i→n(countrefs(i1 ' + ints1)), lastn(i→n(countrefs(i1 ' + ints1)), refs) = r ' + refs1, i→n(countrefs(i1 ' + ints1)) ≤ # refs, rk ∈ st, posints(ints1), 0 ≤ i1, 0 < i0, i2 < i0, 0 ≤ i2, ∀ ty, acc, ints, i, st, rk, refs0. # refs0 < # refs → ( true → ( rk ∈ st ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → rk ∈ addmultiarrays(refs0, i, acc, ty, ints, st))) ⊦ 0 = i0, (~1) = i2, rk ∈ addmultiarrays (r0 ' + firstn(i→n(countrefs(i1 ' + ints1)) * i→n(i2), refs2), i2, r ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(r ' + refs1, i1, [], ty, i1 ' + ints1, st)), 0 ∈ (i1 ' + ints1) .butlast

Rule argument

[speclemmaarg: refs ≠ [] ,r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0 ,[refs, r, refs0] → [refs, r0, refs2] ,T ,list-data ,references ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # [] = 0], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPELIM : refs ≠ [] ⊦ r = refs .first ∧ refs0 = refs .rest ↔ refs = r + refs0])], [EXTRATERMS : (refs, r0, refs2)]]