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Proof step 15 from proof addarray-addaray (previous step)

∃ fs. sk = fs ', 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]), sk ≠ _type ∧ sk ≠ _length

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
0 ≤ i ⊦ (r -- fs ' ∈ addarray(r, ty, i, st) ↔ r -- fs ' ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- fs ']) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- fs ']), fs ' ≠ _type ∧ fs ' ≠ _length

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]