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Proof step 25 from proof addarrays-refin (previous step)

⊦ r -- sk ∈ addarrays(refs, ty, ints, i, st) ↔ r -- sk ∈ st, r ∈ refs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ ints = [] ∨ n→i(# refs) ≠ countrefs(ints) ∨ ¬ posints(ints) → addarrays(refs, ty, ints, i, st) = st
Proof step 24:
¬ (ints = [] ∨ n→i(# refs) ≠ countrefs(ints) ∨ ¬ posints(ints)) ⊦ r -- sk ∈ addarrays(refs, ty, ints, i, st) ↔ r -- sk ∈ st, r ∈ refs

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)]]