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Proof step 22 from proof multi-refin (previous step)

lastn(i→n(countrefs(i1 ' + ints1)), refs) = r + refs1 ∧ lastn(i→n(countrefs(i1 ' + ints1)), refs) ≠ [], # refs = 1 + i→n(countrefs(i1 ' + ints1)) * i→n(i2) + i→n(countrefs(i1 ' + ints1)), rk ∈ st, posints(ints1), 0 ≤ i1, 0 < i0, i2 < i0, 0 ≤ i2, ∀ 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, (~1) = i2, 0 ∈ (i1 ' + ints1) .butlast, rk ∈ addmultiarrays (firstn(# refs - i→n(countrefs(i1 ' + ints1)), refs), i2, r ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(lastn(i→n(countrefs(i1 ' + ints1)), refs), i1, [], ty, i1 ' + ints1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ countrefs(ints) ≠ 0
Proof step 21:
# refs = 1 + i→n(countrefs(i1 ' + ints1)) * i→n(i2) + i→n(countrefs(i1 ' + ints1)), lastn(i→n(countrefs(i1 ' + ints1)), refs) = r ' + refs1, rk ∈ st, posints(ints1), 0 ≤ i1, 0 < i0, i2 < i0, 0 ≤ i2, ∀ 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, (~1) = i2, 0 ∈ (i1 ' + ints1) .butlast, rk ∈ addmultiarrays (firstn(# refs - i→n(countrefs(i1 ' + ints1)), refs), i2, r ' + acc, ty, i0 ' + i1 ' + ints1, addmultiarrays(r ' + refs1, i1, [], ty, i1 ' + ints1, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ r0 ' ↔ r = r0], [CSIMPSEQ : ⊦ r ∈ refs + refs0 ↔ ¬ ¬ (r ∈ refs ∨ r ∈ refs0)], [CSIMPSEQ : ⊦ (refs + refs0) + refs1 = refs + refs0 + refs1], [CSIMPFORWARD : ⊦ r ' + refs = refs0 → r ∈ refs0], [CSIMPSEQ : ⊦ i→n(i) = 0 ↔ i = 0], [CSIMPSEQ : ⊦ lastn(n, refs) = [] ↔ n = 0], [CSIMPSEQ : ⊦ r + refs = r ' + refs])]]