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Proof step 19 from proof addarrays-subset (previous step)

⊦ is_newref_list(refs, st) ∧ 0 ≤ i → (∀ rk. rk ∈ st → rk ∈ addarrays(refs, ty, ints, i, st) ∧ st[rk] = (addarrays(refs, ty, ints, i, st)[rk]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ rk ∈ st → rk ∈ addarrays(refs, ty, ints, i, st)
Proof step 18:
rk ∈ st, is_newref_list(refs, st), 0 ≤ i ⊦ st[rk] = (addarrays(refs, ty, ints, i, st)[rk])

Rule argument

[emptyarg]

Additional information

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