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

¬ # x1 < n + # x1 ⊦ x1 + x0 from n + # x1 = x0 from n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ x1 + x0 from # x1 = x0

Rule argument

Additional information

Used simplifier rules: from-zero-l: ⊦ x from 0 = x;%add-base: ⊦ n + 0 = n; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%le-add: ⊦ m + m0 ≤ m ↔ m0 = 0; from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat