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Proof step 10 from proof getarray-in-01 (previous step)

0 < i3, ∀ st, i0, r, val, i. i→n(i) < i→n(i3 + 1) → (val ∈ getarray(r, i0, i, st) ∧ 0 < i → (∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i ≤ i2 ∧ st[r -- i2 '] = val)) ⊦ val ≠ st[r -- (i0 + i3) '], ∃ i. ¬ i < i0 ∧ ¬ i0 + i3 < i ∧ st[r -- i '] = val

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
0 < i3, ∀ st, i0, r, val, i. i→n(i) < i→n(i3 + 1) → (val ∈ getarray(r, i0, i, st) ∧ 0 < i → (∃ i2. ¬ i2 < i0 ∧ ¬ i0 + i ≤ i2 ∧ st[r -- i2 '] = val)) ⊦ ∃ i. ¬ i < i0 ∧ ¬ i0 + i3 < i ∧ st[r -- i '] = st[r -- (i0 + i3) ']

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]