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Proof step 10 from proof split-first (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
n = m + 1, # y = m, 1 ≤ n ⊦ n = 0, ∃ y0, z0. a ' + y + z = y0 + z0 ∧ # y0 = n

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m