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Proof step 2 from proof split-first (previous step)

∀ n. n ≤ # x0 → (∃ y, z. x0 = y + z ∧ # y = n), n ≤ #(a + x0) ⊦ ∃ y, z. a + x0 = y + z ∧ # y = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ # x0 + 1, ∀ n. n ≤ # x0 → (∃ y, z. x0 = y + z ∧ # y = n) ⊦ ∃ y, z. a ' + x0 = y + z ∧ # y = n

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x;%len-05: ⊦ #(a ' + x) = (# x)+1;%a: ⊦ (x + y) + z = x + y + z;%succ: ⊦ m +1 = m + 1; from specification nat-basic2