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Proof step 61 from proof putarray-addarray-override (previous step)

0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- sk ∈ addarray(r, ty, vals0, st) ∨ r0 = r ∧ is_indexkey(sk) ∧ i ≤ sk .index ∧ sk .index < i + n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

User applied rule cut formula

And obtained the following new goals

Proof step 31:
is_indexkey(sk), 0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- sk ∈ addarray(r, ty, vals0, st) ∨ r0 = r ∧ is_indexkey(sk) ∧ i ≤ sk .index ∧ sk .index < i + n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)
Proof step 60:
0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ is_indexkey(sk), r0 -- sk ∈ addarray(r, ty, vals0, st) ∨ r0 = r ∧ is_indexkey(sk) ∧ i ≤ sk .index ∧ sk .index < i + n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

Rule argument

[fma-arg: is_indexkey(sk)]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRAFMAS : (is_indexkey(sk))]]