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Proof step 5 from proof from-from (previous step)

n + m ≤ # x ⊦ x from m from n = x from m + n

User applied rule structural induction

And obtained the following new goals

Proof step 6:
n + m ≤ #(a + x0), ∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n ⊦ a + x0 from m from n = a + x0 from m + n

Rule argument

indvar: x

Additional information

Used simplifier rules: from-zero-l: ⊦ x from 0 = x;%list freely generated by [], +;;% ⊦ # [] = 0; (s)% ⊦ m ≤ 0 ↔ m = 0; (s)% ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; (ws)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic