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Proof step 6 from proof less (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 7:
n + n0 = m + m0, n < m, n0 < m0 → n + n0 < m + m0 ⊦ m = 0, m0 ≠ n0 ∧ ¬ n0 < m0

Rule argument

pos: right 3

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m;%ls-04: ⊦ m < n ↔ ¬(m = n ∨ n < m);%f: ⊦ m < n → n ≠ 0;%l: ⊦ m + n < m + n0 ↔ n < n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;