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Proof step 62 from proof multi-at-other (previous step)

# (i0 ' + ints0) = 1, # refs = i→n(countrefs(i ' + ints0)), i ≤ i0, 0 ≤ i, posints(ints0), 0 ≤ i0, ∀ sk, 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 → (addmultiarrays(refs0, i, acc, ty, ints, st)[r -- sk]) = st[r -- sk])) ⊦ (addarray(refs .first, ty, (i0 ' + ints0) .first, st)[r -- sk]) = st[r -- sk], r ∈ refs, 0 ∈ (i0 ' + ints0) .butlast

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 60:
⊦ countrefs(i ') = 1
Proof step 61:
# refs = 1, 0 ≤ i0, 0 ≤ i, i ≤ i0, ∀ sk, st, ty, acc, ints, i, r, refs0. # refs0 < 1 → ( true → ( ¬ r ∈ refs0 ∧ # refs0 = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → (addmultiarrays(refs0, i, acc, ty, ints, st)[r -- sk]) = st[r -- sk])) ⊦ (addarray(refs .first, ty, i0, st)[r -- sk]) = st[r -- sk], r ∈ refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i ∈ []], [CSIMPSEQ : ⊦ i ' .butlast = []], [CSIMPSEQ : ⊦ posints([])], [CSIMPSEQ : ⊦ ints + [] = ints], [CSIMPSEQ : ⊦ (i ' + ints) .first = i], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ 1 = m + 1 ↔ m = 0], [CSIMPSEQ : ⊦ (ints + ints0) + ints1 = ints + ints0 + ints1], [CSIMPSEQ : ⊦ # i ' = 0 +1], [CSIMPSEQ : ⊦ # (ints + ints0) = # ints + # ints0])]]