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Proof step 47 from proof addarrays-at-other (conclusion)

⊦ ¬ r ∈ refs ∧ # refs = i→n(countrefs(ints)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) → (addarrays(refs, ty, ints, i, st)[r -- sk]) = st[r -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 46:
# refs = i→n(countrefs(ints)), posints(ints), 0 ≤ i ⊦ (addarrays(refs, ty, ints, i, st)[r -- sk]) = st[r -- sk], ints = [], r ∈ refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j])]]