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Proof step 59 from proof getarray-subset (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ rk ∈ st ∧ st ⊆ st1 → st[rk] = st1[rk]
Proof step 10:
⊦ ¬ r -- _length ∈ st → ¬ is_array(r, st)
Proof step 58:
r -- _length ∈ st ∧ st ⊆ st0, is_array(r, st), st ⊆ st0, 0 ≤ st0[r -- _length] .intval, 0 ≤ st0[r -- _length] .intval ⊦ st0[r -- _length] .intval = st0[r -- _length] .intval ∧ (n < i→n(st0[r -- _length] .intval) → st[r -- n→i(n) '] = (getarray(r, st0)[n]))

Rule argument

[speclemmaarg: ,rk ∈ st ∧ st ⊆ st1 → st[rk] = st1[rk] ,[rk, st, st1] → [r -- _length, st, st0] ,F ,store ,javastore ,subset-elem-eq ]

Additional information

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