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Proof step 24 from proof putarray-addarray (previous step)

0 ≤ i ∧ 0 ≤ j ∧ i ≤ j + n→i(# vals) ∧ j ≤ i ∧ 0 ≤ j + n→i(# vals), ∀ st, val, r, j, i, vals0. # vals0 < # vals → ( 0 ≤ i ∧ 0 ≤ j ∧ i ≤ j + n→i(# vals0) ∧ j ≤ i ∧ 0 ≤ j + n→i(# vals0) → putArray(vals0, r, j, addarray(r, val, i, st)) = addarray(r, mklist(val, i→n(j)) + vals0, 0, st)) ⊦ putArray(vals, r, j, addarray(r, val, i, st)) = addarray(r, mklist(val, i→n(j)) + vals, 0, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
0 ≤ j + n→i(# vals), j ≤ i, i ≤ j + n→i(# vals), 0 ≤ j, 0 ≤ i, ∀ st, val, r, j, i, vals0. # vals0 < # vals → ( 0 ≤ i ∧ 0 ≤ j ∧ i ≤ j + n→i(# vals0) ∧ j ≤ i ∧ 0 ≤ j + n→i(# vals0) → putArray(vals0, r, j, addarray(r, val, i, st)) = addarray(r, mklist(val, i→n(j)) + vals0, 0, st)) ⊦ putArray(vals, r, j, addarray(r, val, i, st)) = addarray(r, mklist(val, i→n(j)) + vals, 0, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]