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Proof step 52 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) ⊦ r -- n→i(n) ' ∈ st, st0[r -- _length] .intval = 0, st[r -- n→i(n) '] = st0[r -- n→i(n) '], getarray(r, st0) = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 51:
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) ⊦ getarray(r, st0) = [], st0[r -- _length] .intval = 0, noval = st0[r -- n→i(n) '], r -- n→i(n) ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ∅[rk] = noval], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]