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Proof step 5 from proof two-loup-02 (previous step)

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

User applied rule insert elim lemma

And obtained the following new goals

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

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: z: ⊦ 0 + m = m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%twols: ⊦ m < n → (m + n < 2 ↔ m = 0 ∧ n = 1); from specification nat-basic2%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m0, m1