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Proof step 2 from proof put-app-right (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 3:
n = n0 + # x, # x ≤ n ⊦ (x + y)[n, a] = x + y[n0, a]

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, # x, n0] flags:

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # x, n0