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Proof step 22 from proof splice-resize-less (previous step)

n < n1 ⊦ (max(# ar1, n1 + m0) ≤ n ⊃ zero;splice(ar0, n0, ar1, n1, m0)[n]) = (# ar1 ≤ n ⊃ zero;ar1[n]), m0 = 0

User applied rule case distinction

And obtained the following new goals

Proof step 23:
n < n1 ⊦ splice(ar0, n0, ar1, n1, m0)[n] = (# ar1 ≤ n ⊃ zero;ar1[n]), m0 = 0, max(# ar1, n1 + m0) ≤ n

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)% ⊦ ¬ max(n, m) < n; (s)