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Proof step 201 from proof putarray-addarray-01 (previous step)

0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ sk = _length, (r -- sk ∈ putArray(vals, r, i, addarray(r, ty, j, st)) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)) ∧ (putArray(vals, r, i, addarray(r, ty, j, st))[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- sk]), ¬ is_indexkey(sk) ∧ sk ≠ _type

User applied rule case distinction

And obtained the following new goals

Proof step 151:
0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ r -- sk ∈ putArray(vals, r, i, addarray(r, ty, j, st)) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st), sk = _length, ¬ is_indexkey(sk) ∧ sk ≠ _type
Proof step 200:
0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ (putArray(vals, r, i, addarray(r, ty, j, st))[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- sk]), sk = _length, ¬ is_indexkey(sk) ∧ sk ≠ _type

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ()]]