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

x0 = y + z ∧ # y = n - 1, n ≤ # x0 + 1 ⊦ ∃ 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 ≤ ((n - 1) + # z) + 1 ⊦ ∃ 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;%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2