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Proof step 64 from proof getarray-subset (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ vals = vals0 ↔ # vals = # vals0 ∧ (∀ n. n < # vals → vals[n] = vals0[n])
Proof step 63:
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]))

Rule argument

[speclemmaarg: , vals = vals0 ↔ # vals = # vals0 ∧ (∀ n. n < # vals → vals[n] = vals0[n]) ,[vals0, vals] → [getarray(r, st0), getarray(r, st)] ,F ,list-del ,javavalues ,extension ]

Additional information

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