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Proof step 49 from proof splice-resize-less (previous step)

n < n2 + n1, n < # ar1, n1 + m0 ≤ # ar1, n2 ≤ m0, n1 + n2 ≤ # ar1 ⊦ (n < n1 ∨ n1 + m0 ≤ n ⊃ ar1[n];ar0[n + n0 - n1]) = (n < n1 ⊃ ar1[n];ar0[n + n0 - n1]), m0 = 0, n2 = 0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic