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Proof step 5 from proof copy-new (previous step)

true, ∀ j0, r1, r0, r, i, i0, st. i→n(j0) < i→n(j) → (true → (r1 ≠ r0 → (newref(r1, arraycopy(r, i, r0, i0, j0, st)) ↔ newref(r1, st)))) ⊦ r1 ≠ r0 → (newref(r1, arraycopy(r, i, r0, i0, j, st)) ↔ newref(r1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ j0, r1, r0, r, i, i0, st. i→n(j0) < i→n(j) → (true → (r1 ≠ r0 → (newref(r1, arraycopy(r, i, r0, i0, j0, st)) ↔ newref(r1, st)))) ⊦ r1 = r0, newref(r1, arraycopy(r, i, r0, i0, j, st)) ↔ newref(r1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]