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Proof step 65 from proof multg-at-ref (previous step)

¬ (# ints > 1 ∧ ints .first ≤ 0), i→n(countrefs2(ints)) ≤ # refs, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(ints) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ # ints = 1, (addarraymultlist(r, ty, st, refs, ints, ints0, i)[r0 -- sk]) = st[r0 -- sk], r = r0, r0 ∈ refs

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 64:
i→n(countrefs2(ints)) ≤ # refs, ∀ sk, i, ints0, st, ty, r0, r, refs, ints1. ints#(ints1) < ints#(ints) → (i→n(countrefs2(ints1)) ≤ # refs → r = r0 ∨ (addarraymultlist(r, ty, st, refs, ints1, ints0, i)[r0 -- sk]) = st[r0 -- sk] ∨ r0 ∈ refs) ⊦ # ints = 1, (addarraymultlist(r, ty, st, refs, ints, ints0, i)[r0 -- sk]) = st[r0 -- sk], r = r0, r0 ∈ refs, ints ≠ [] ∧ ¬ 0 < ints .first

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ # ints = 0 ↔ ints = []], [CSIMPSEQ : ⊦ 1 < m ↔ ¬ (m = 0 ∨ m = 1)], [CSIMPSEQ : ⊦ m > n ↔ n < m])]]