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Proof step 8 from proof getarray-subset (previous step)

¬ 0 ≤ st0[r -- _length] .intval, 0 ≤ st[r -- _length] .intval, is_array(r, st), st ⊆ st0 ⊦ i→n(st[r -- _length] .intval) = # getarray(r, st0) ∧ (n < i→n(st[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 7:
st0[r -- _length] .intval < 0, 0 ≤ st[r -- _length] .intval, st ⊆ st0, is_array(r, st) ⊦ i→n(st[r -- _length] .intval) = # getarray(r, st0) ∧ (n < i→n(st[r -- _length] .intval) → st[r -- n→i(n) '] = (getarray(r, st0)[n]))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < 0 ∧ 0 ≤ j → i < j], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPFORWARD : ⊦ j < i ∧ i ≤ 0 → j < 0], [CSIMPFORWARD : ⊦ i < j ∧ j ≤ j0 → i < j0], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i])]]