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Proof step 63 from proof getarray-subset (previous step)

is_array(r, st), st ⊆ st0 ⊦ # getarray(r, st) = # getarray(r, st0) ∧ (∀ n. n < # getarray(r, st) → (getarray(r, st)[n]) = (getarray(r, st0)[n]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 62:
is_array(r, st), st ⊆ st0 ⊦ # getarray(r, st) = # getarray(r, st0) ∧ (n < # getarray(r, st) → st[r -- n→i(n) '] = (getarray(r, st0)[n]))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n < # getarray(r, st) → (getarray(r, st)[n]) = st[r -- n→i(n) ']])]]