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Proof step 5 from proof two-loup-04 (previous step)

m + n = m0 * 2 + n0, n0 < 2, 2 + m < m0, m + 1 < n, m + 1 < m0, m < n, n + 1 < m0 ⊦ n = 0, m0 - 1 = 0, m + 1 = n, m0 = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
m0 = m1 + 1, m + n = m0 * 2 + n0, n0 < 2, 2 + m < m0, m + 1 < n, m + 1 < m0, m < n, n + 1 < m0 ⊦ n = 0, m1 = 0, m + 1 = n, m0 = 0

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [m0, m1] flags:

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m0, m1