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Proof step 14 from proof getarray-subset (previous step)

¬ 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
st ⊆ st0, is_array(r, st), r -- _length ∈ st, r ∈ st, n < i→n(st0[r -- _length] .intval), 0 < st0[r -- _length] .intval, # getarray(r, st0) ≤ n, # getarray(r, st0) < i→n(st0[r -- _length] .intval) ⊦ st[r -- n→i(n) '] = (getarray(r, st0)[n]), st0[r -- _length] .intval = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ m ≤ n ∧ n < n0 → m < n0], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m])]]