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Proof step 14 from proof addarrays-subset (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ ints = [] ∨ n→i(# refs) ≠ countrefs(ints) ∨ ¬ posints(ints) → addarrays(refs, ty, ints, i, st) = st
Proof step 13:
¬ (ints = [] ∨ n→i(# refs) ≠ countrefs(ints) ∨ ¬ posints(ints)), r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk]), ¬ r ∈ refs ∧ # refs = i→n(countrefs(ints)) ∧ ints ≠ [] ∧ posints(ints)

Rule argument

[crewrite-arg: addarrays-empty , F , [st, i, ints, ty, refs] → [st, i, ints, ty, refs] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, i, ints, ty, refs)]]