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Proof step 57 from proof getarray-subset (previous step)

st ⊆ st0, is_array(r, st), r -- _length ∈ st, r ∈ st, n < i→n(st0[r -- _length] .intval), 0 < st0[r -- _length] .intval ⊦ st[r -- n→i(n) '] = (getarray(r, st0)[n]), st0[r -- _length] .intval = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ n < # getarray(r, st) → (getarray(r, st)[n]) = st[r -- n→i(n) ']
Proof step 14:
¬ n < # getarray(r, st0), st ⊆ st0, is_array(r, st), r -- _length ∈ st, r ∈ st, n < i→n(st0[r -- _length] .intval), 0 < st0[r -- _length] .intval ⊦ st[r -- n→i(n) '] = (getarray(r, st0)[n]), st0[r -- _length] .intval = 0
Proof step 56:
n < # getarray(r, st0), st ⊆ st0, is_array(r, st), r -- _length ∈ st, r ∈ st, n < i→n(st0[r -- _length] .intval), 0 < st0[r -- _length] .intval ⊦ st[r -- n→i(n) '] = st0[r -- n→i(n) '], st0[r -- _length] .intval = 0

Rule argument

[speclemmaarg: ,n < # getarray(r, st) → (getarray(r, st)[n]) = st[r -- n→i(n) '] ,[n, st, r] → [n, st0, r] ,F ,storearrays , ,getarray-get ]

Additional information

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