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

∀ x0, a, y. n0 ≤ # x0 + 1 ∧ (∀ n, y. n ≤ # x0 → x0 + y from n = (x0 from n) + y) → (a ' + x0) + y from n0 = (a ' + x0 from n0) + y, n0 +1 ≤ # x0 + 1, ∀ n, y. n ≤ # x0 → x0 + y from n = (x0 from n) + y ⊦ (a ' + x0) + y from n0 +1 = (a ' + x0 from n0 +1) + y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 ≤ # x0, ∀ x0, a, y. n0 ≤ # x0 + 1 ∧ (∀ n, y. n ≤ # x0 → x0 + y from n = (x0 from n) + y) → (a ' + x0) + y from n0 = (a ' + x0 from n0) + y, ∀ n, y. n ≤ # x0 → x0 + y from n = (x0 from n) + y ⊦ x0 + y from n0 = (x0 from n0) + y

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