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Proof step 5 from proof two-loup-09 (previous step)

m + n = m0 * 2 + n0, n0 < 2 ⊦ m0 ≠ m ↔ m ≠ n ∧ m + 1 ≠ n

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: numeral-19: m2 ≠ 0 ⊦ (* m2) * m + n = m + n0 ↔ (m2 - 1) * m + n = n0; from specification nat-mult%one: ⊦ m0 * 1 = m0; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%twols-02: ⊦ m ≠ 0 → (m < 2 ↔ m = 1); from specification nat-basic1%two-02: ⊦ m + m = 2 * m; from specification nat-mult%modeq-01: ⊦ n0 < n → (m * n + n0 = m0 * n ↔ m = m0 ∧ n0 = 0); from specification nat-mult%modeq: ⊦ n0 < n ∧ n1 < n → (m * n + n0 = m0 * n + n1 ↔ m = m0 ∧ n0 = n1); from specification nat-mult