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Proof step 23 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n, ar = subarray(ar, 0, n) + subarray(ar, n, n1 + m), n1 ≠ 0 → subarray(ar, n, n1 + m) = subarray(subarray(ar, n, n1 + m), 0, m) + subarray(subarray(ar, n, n1 + m), m, n1) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

User applied rule case distinction

And obtained the following new goals

Proof step 24:
# ar = (n1 + m) + n, ar = subarray(ar, 0, n) + subarray(ar, n, n1 + m), n1 = 0 ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)
Proof step 35:
# ar = (n1 + m) + n, ar = subarray(ar, 0, n) + subarray(ar, n, n1 + m), subarray(ar, n, n1 + m) = subarray(subarray(ar, n, n1 + m), 0, m) + subarray(subarray(ar, n, n1 + m), m, n1) ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m)

Rule argument

pos: left 3

Proof step 23 from proof subarray-eq-elim (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information