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Proof step 3 from proof less-01 (previous step)

n0 + m < m0 ⊦ m0 = 0, n + m ≠ m0 ∧ ¬ m0 < n + m, n0 ≠ n ∧ ¬ n < n0

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: l: ⊦ m + n < m + n0 ↔ n < n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%ls-02: ⊦ m < n → ¬ n < m;%f: ⊦ m < n → n ≠ 0;%trans-01: ⊦ n < n0 ∧ n0 < n1 → n +1 < n1;%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1;%succ: ⊦ m +1 = m + 1;%l: ⊦ m + n < m + n0 ↔ n < n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%ls-02: ⊦ m < n → ¬ n < m;