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Proof step 67 from proof multg-at-ref (previous step)

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 = 1 ∧ ¬ 0 < ints .first, # ints = 1 ∧ ¬ ints .first ≤ 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 66:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]