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Proof step 11 from proof q-add-01 (previous step)

m0 = n ∨ ¬ m0 < n ⊦ (m0 + 1) + 0 -1 = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 2

Additional information

Used simplifier rules: eq-01: ⊦ m + n = m + n0 ↔ n = n0;%one-06: ⊦ 1 = m + 1 ↔ m = 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%eq-02: ⊦ m = m + n ↔ n = 0;%s-04: ⊦ m < n + 1 ↔ ¬ n < m;%ref: ⊦ ¬ n < n;%f: ⊦ m < n → n ≠ 0;%eq-02: ⊦ m = m + n ↔ n = 0;%oz: ⊦ m + 1 ≠ 0;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;%l-02: ⊦ ¬ m + n < m;%nls-fst: ⊦ ¬ m < n → ¬ m + m0 < n;