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Proof step 71 from proof multi-isarrayref (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 70:
n→i(# acc) + i = ints .first, # refs = i→n(countrefs(i ' + ints .rest)), r ∈ refs, posints(ints), 0 ≤ i, ∀ st, ty, ints, i, acc, r, refs0. # refs0 < # refs → ( true → ( r ∈ refs0 ∧ n→i(# acc) + i = ints .first ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → is_arrayref(r, (addmultiarrays(refs0, i, acc, ty, ints, st)[r -- _length]) .intval, addmultiarrays(refs0, i, acc, ty, ints, st)))) ⊦ ints = [], is_arrayref(r, (addmultiarrays(refs, i, acc, ty, ints, st)[r -- _length]) .intval, addmultiarrays(refs, i, acc, ty, ints, st)), 0 ∈ ints .butlast

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ i = j + n→i(n) → j ≤ i], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0])]]