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Proof step 6 from proof to-plus-from-fin (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 ≤ # x0, ∀ x0, a. n0 ≤ 1 + # x0 ∧ (∀ n. n ≤ # x0 → (x0 to n) + (x0 from n) = x0) → (a ' + x0 to n0) + (a ' + x0 from n0) = a ' + x0, ∀ n. n ≤ # x0 → (x0 to n) + (x0 from n) = x0 ⊦ (x0 to n0) + (x0 from n0) = x0

Rule argument

Additional information

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