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Proof step 3 from proof less-02 (previous step)

n0 + n1 + n < m ⊦ m = 0, n0 + n1 + n2 ≠ m ∧ ¬ m < n0 + n1 + n2, n ≠ n2 ∧ ¬ n2 < n

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

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