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Proof step 5 from proof split-rest (previous step)

# y = n0, # y + # z = n0 + n, n ≤ # y + # z ⊦ ∃ y0, z0. y + z = y0 + z0 ∧ # z0 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ ∃ y0, z0. y + z = y0 + z0 ∧ # z0 = # z

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)