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Proof step 8 from proof less (previous step)

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

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 3

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%ref: ⊦ ¬ n < n;%ls-02: ⊦ m < n → ¬ n < m;