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Proof step 2 from proof getarray-superset-01 (previous step)

is_array(r, st), st ⊆ st0 ⊦ getarray(r, st0) = getarray(r, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st ⊆ st0 ∧ is_array(r, st) → getarray(r, st) = getarray(r, st0)

Rule argument

[crewrite-arg: getarray-subset, F, [st, st0, r] → [st, st0, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, st0, r)]]