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

n = n0 + # x ∧ # x ≤ n, # x < n ⊦ x + y from n = y from n0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ x + y from n0 + # x = y from n0, n0 + # x = 0, n0 = 0

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat