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Proof step 32 from proof fill-append-shift (previous step)

n < # ar0 + # ar1, n < # ar0 ⊦ (n0 + # ar0 ≤ n ∧ n < (n0 + # ar0) + m ⊃ a;ar0[n]) = ar0[n]

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: 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)% ⊦ ¬ m + n < m; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic