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Proof step 9 from proof split-first (previous step)

1 ≤ n, # y = n - 1, n ≤ # z + n ⊦ ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
# y = n - 1, 1 ≤ n ⊦ n = 0, ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

Rule argument

Additional information

Used simplifier rules: not-zero: 0 < n ⊦ (* n) ≤ m → m ≠ 0; (forward) from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat