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Proof step 49 from proof subarray-eq-elim (previous step)

# ar = (n1 + m) + n ⊦ 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 50:
# ar = (n1 + m) + n, subarray(ar, n, m) = ar1 ⊦ ∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m
Proof step 61:
# ar = (n1 + m) + n, ∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = # ar1 + # ar2 - m ⊦ subarray(ar, n, m) = ar1

Rule argument

pos: right 1

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