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Proof step 8 from proof prefixn-eq-prefixn-app (previous step)

x0 = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1) ⊦ n0 ≤ # x0, n1 < # x0 ∧ n0 = n1, # x0 ≤ n0 ∧ # x0 ≤ n1 ∧ x1 = [], # x0 ≤ n0 ∧ n1 = # x0 ∧ x1 ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
x0 = (n1 ≤ # x0 + # x1 ⊃ firstn(n1, x0 + x1);x0 + x1), # x0 < n0 ⊦ # x0 ≤ n1 ∧ x1 = [], n1 = # x0 ∧ x1 ≠ []

Rule argument

Additional information

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