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Proof step 3 from proof append-mod (previous step)

n = n0 + # x, n < # x + # x0, # x ≤ n ⊦ # x + # x0 = 0, (x + x0)[n, a] = x + x1 ↔ x0[n0, a] = x1, x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n0 < # x0 ⊦ x0 = [], (x + x0)[n0 + # x, a] = x + x1 ↔ x0[n0, a] = x1, # x + # x0 = 0

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lep: ⊦ m ≤ m + n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2