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Proof step 67 from proof multi-lengthin (previous step)

ints = i0 + ints0, # refs = i→n(countrefs(i ' + ints0)), r ∈ refs, posints(ints), i ≤ i0, 0 ≤ i, 0 ≤ i0, ∀ st, ty, acc, ints, i, r, refs0. # refs0 < # refs → ( 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)) ⊦ ints = [], 0 ∈ ints .butlast, r -- _length ∈ addmultiarrays(refs, i, acc, ty, ints, st) ∧ is_integervalue(addmultiarrays(refs, i, acc, ty, ints, st)[r -- _length]) ∧ ¬ (addmultiarrays(refs, i, acc, ty, ints, st)[r -- _length]) .intval < 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
# refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, r ∈ refs, posints(ints0), 0 ≤ i0, ∀ st, ty, acc, ints, i, r, refs0. # refs0 < # refs → ( 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)) ⊦ 0 ∈ (i0 ' + ints0) .butlast, r -- _length ∈ addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st) ∧ is_integervalue(addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)[r -- _length]) ∧ ¬ (addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)[r -- _length]) .intval < 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j], [CSIMPSEQ : ⊦ posints(i ' + ints) ↔ 0 ≤ i ∧ posints(ints)], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ i ' + ints ≠ []], [CSIMPSEQ : ⊦ i + ints = i ' + ints])]]