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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ 1 + # x0, ∀ 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: ⊦ #(x + y) = # x + # y;%len-02: ⊦ # a ' = 0 +1;%a: ⊦ (x + y) + z = x + y + z;