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Proof step 7 from proof from-from (previous step)

∀ n, x0, a. n + m0 ≤ # x0 + 1 ∧ (∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n) → a ' + x0 from m0 from n = a ' + x0 from m0 + n, n + m0 +1 ≤ # x0 + 1, ∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n ⊦ a ' + x0 from m0 +1 from n = a ' + x0 from m0 +1 + n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
n + m0 ≤ # x0, ∀ n, x0, a. n + m0 ≤ # x0 + 1 ∧ (∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n) → a ' + x0 from m0 from n = a ' + x0 from m0 + n, ∀ n, m. n + m ≤ # x0 → x0 from m from n = x0 from m + n ⊦ x0 from m0 from n = x0 from m0 + n

Rule argument

Additional information

Used simplifier rules: from-succ-cons-l-lem: ⊦ a ' + x from n + 1 = x from n;%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2