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Proof step 17 from proof multi-lengthin (previous step)

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

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[speclemmaarg: n ≤ m ,n0 = m - n ↔ m = n0 + n ,[m, n, n0] → [# refs2, 1 + # refs1, n] ,T ,nat , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m < n → (n ≤ m ↔ false)], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m * n = n * m], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ m + 1 ≤ n ↔ m < n], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPFORWARD : ⊦ m = n + n0 ∧ m = n + n1 → n0 = n1], [CSIMPFORWARD : ⊦ m < n ∧ n = n0 → m < n0], [CSIMPSEQ : ⊦ m + n = m + n0 ↔ n = n0], [CSIMPSEQ : ⊦ i ≤ j → (¬ ¬ j < i ↔ false)], [CSIMPELIM : n ≤ m ⊦ n0 = m - n ↔ m = n0 + n])], [EXTRATERMS : (# refs2, 1 + # refs1, n)]]