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Proof step 6 from proof split-first (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# y = n - 1, n ≤ 1 + (n - 1) + # z ⊦ ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

Rule argument

Additional information

Used simplifier rules: len: ⊦ #(x + y) = # x + # y;%a: ⊦ (x + y) + z = x + y + z;