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Proof step 4 from proof isprefix-prefixn (previous step)

x0 ⊑ x1, n = # x0 ∨ # x1 < n ∧ x0 = x1 ⊦ x0 = (n ≤ # x1 ⊃ firstn(n, x1);x1)

Heuristic if-then-else split applied rule case distinction

And closed the goal!

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%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ # y < # x → ¬ x ⊑ y; (s)% ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ # x0 = n ∧ x0 ⊑ x2 → firstn(n, x2) = x0; (s)