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Proof step 2 from proof rangeeq-subarray-5 (previous step)

range=(avol, 5 + n, avol0, 5 + n, m0) ⊦ subarray(avol, 5 + n, m0) = subarray(avol0, 5 + n, m0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ avol = avol0 ↔ # avol = # avol0 ∧ (∀ n. n < # avol → avol[n] = avol0[n])
Proof step 4:
range=(avol, 5 + n, avol0, 5 + n, m0) ⊦ # subarray(avol, 5 + n, m0) = # subarray(avol0, 5 + n, m0) ∧ (∀ n0. n0 < # subarray(avol, 5 + n, m0) → subarray(avol, 5 + n, m0)[n0] = subarray(avol0, 5 + n, m0)[n0])

Rule argument

spec: array, inst: avolume name: Extension seq: ⊦ avol = avol0 ↔ # avol = # avol0 ∧ (∀ n. n < # avol → avol[n] = avol0[n]) subst: [avol0, avol] → [subarray(avol0, 5 + n, m0), subarray(avol, 5 + n, m0)] flags: <all paths>

Additional information

Used terms: subarray(avol0, 5 + n, m0), subarray(avol, 5 + n, m0)