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Proof step 58 from proof getarray-subset (previous step)

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]))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPSEQ : ⊦ i ≠ j → (i ≤ j ↔ i < j)], [CSIMPSEQ : ⊦ i→n(i) = 0 ↔ i = 0], [CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st], [CSIMPFORWARD : ⊦ m < n → n ≠ 0])]]