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Proof step 3 from proof put-app-right (previous step)

n = n0 + # x, # x ≤ n ⊦ (x + y)[n, a] = x + y[n0, a]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (x + y)[n0 + # x, a] = x + y[n0, a]

Rule argument

Additional information

Used simplifier rules: 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