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Proof step 25 from proof multi-refin (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 24:
i = i2 + 1, # refs = i→n(countrefs(i ' + i1 ' + ints1)), 0 < i0, i ≤ i0, 0 < i, rk ∈ st, posints(ints1), 0 ≤ i1, ∀ 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, i = 0, 0 ∈ (i1 ' + ints1) .butlast, rk ∈ addmultiarrays (firstn(# refs - i→n(countrefs(i1 ' + ints1)), refs), i2, lastn(i→n(countrefs(i1 ' + ints1)), refs) .first ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(lastn(i→n(countrefs(i1 ' + ints1)), refs), i1, [], ty, i1 ' + ints1, st))

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i, 1, i2] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i, 1, i2)]]