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Proof step 55 from proof getarray-subset (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 15:
⊦ rk ∈ st ∧ st ⊆ st1 → st[rk] = st1[rk]
Proof step 54:
¬ (r -- n→i(n) ' ∈ st ∧ st ⊆ st0), 0 < st0[r -- _length] .intval, n < i→n(st0[r -- _length] .intval), r ∈ st, r -- _length ∈ st, is_array(r, st), st ⊆ st0, n < # getarray(r, st0) ⊦ st0[r -- _length] .intval = 0, st[r -- n→i(n) '] = st0[r -- n→i(n) '], getarray(r, st0) = []

Rule argument

[speclemmaarg: ,rk ∈ st ∧ st ⊆ st1 → st[rk] = st1[rk] ,[rk, st, st1] → [r -- n→i(n) ', st, st0] ,F ,store ,javastore ,subset-elem-eq ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r -- n→i(n) ', st, st0)]]