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Proof step 11 from proof split-first (previous step)

# y = m, n = m + 1, 1 ≤ n ⊦ n = 0, ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = # y + 1

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 + 1 ≠ 0; (s)