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Proof step 4 from proof split-first (previous step)

n ≤ 1 + # x0, n -1 ≤ # x0 → (∃ y, z. x0 = y + z ∧ # y = n -1) ⊦ ∃ y, z. a ' + x0 = y + z ∧ # y = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n ≤ 1 + # x0, n - 1 ≤ # x0 → x0 = y + z ∧ # y = n - 1 ⊦ ∃ y, z. a ' + x0 = y + z ∧ # y = n

Rule argument

Additional information

Used simplifier rules: ⊦ n -1 = n - 1; (s)