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Proof step 27 from proof addarrays-out (conclusion)

⊦ (addarrays(refs, ty, ints, i, st)[_out]) = st[_out]

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 26:
¬ (ints = [] ∨ n→i(# refs) ≠ countrefs(ints) ∨ ¬ posints(ints)) ⊦ (addarrays(refs, ty, ints, i, st)[_out]) = st[_out]

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)]]