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Proof step 2 from proof append-mod (previous step)

n < # x + # x0, # x ≤ n ⊦ (x + x0)[n, a] = x + x1 ↔ x0[n - # x, a] = x1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n < # x + # x0, # x ≤ n ⊦ (x + x0)[n, a] = x + x1 ↔ x0[n - # x, a] = x1, x0 = []

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # x = 0 ↔ x = []; (s)% ⊦ m < m + n ↔ n ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic