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Proof step 54 from proof addarray-addaray (previous step)

0 ≤ i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 ∨ r = r0 ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r = r0 ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ j ≤ sk .index) → (addarray(r, ty, j, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 30:
¬ (r ≠ r ∨ r = r ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r = r ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ i ≤ sk .index)), 0 ≤ i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk])
Proof step 53:
r ≠ r ∨ r = r ∧ is_fskey(sk) ∧ sk ≠ _length ∧ sk ≠ _type ∨ r = r ∧ is_indexkey(sk) ∧ (sk .index < 0 ∨ i ≤ sk .index), 0 ≤ i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ st[r -- sk] = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk])

Rule argument

[crewrite-arg: at-1dim-other , F , [sk, r0, st, j, ty, r] → [sk, r, st, i, ty, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r, st, i, ty, r)]]