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Proof step 3 from proof split-first (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 4:
n ≤ # x0 + 1, n -1 ≤ # x0 → (∃ y, z. x0 = y + z ∧ # y = n -1) ⊦ ∃ y, z. a ' + x0 = y + z ∧ # y = n

Rule argument

pos: left 2 terms: n -1 abort: F discard: T computed: F

Additional information

Used terms: n -1