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Proof step 4 from proof split-rest (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
# y + # z = n0 + n, # y = n0, n ≤ # y + # z ⊦ ∃ y0, z0. y + z = y0 + z0 ∧ # z0 = n

Rule argument

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

Additional information

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