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Proof step 60 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) = i→n(st0[r -- _length] .intval) ∧ (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 59:
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]))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ 0 ≤ i → abs(i) = i], [CSIMPSEQ : ⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j)])]]