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Proof step 16 from proof getarray-addarray-05 (previous step)

i ≤ j ∧ 0 ≤ i ∧ 0 ≤ j, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (i ≤ j0 ∧ 0 ≤ i ∧ 0 ≤ j0 → getarray(r, 0, i, addarray(r, val, j0, st)) = mklist(val, i→n(i))) ⊦ getarray(r, 0, i, addarray(r, val, j, st)) = mklist(val, i→n(i))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
0 ≤ j, 0 ≤ i, i ≤ j, ∀ st, val, r, i, j0. i→n(j0) < i→n(j) → (i ≤ j0 ∧ 0 ≤ i ∧ 0 ≤ j0 → getarray(r, 0, i, addarray(r, val, j0, st)) = mklist(val, i→n(i))) ⊦ getarray(r, 0, i, addarray(r, val, j, st)) = mklist(val, i→n(i))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]