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Proof step 10 from proof q-add-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
m0 = n ∨ ¬ m0 < n ⊦ (m0 + 1) + 0 -1 = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

Rule argument

Additional information

Used simplifier rules: z-02: ⊦ 0 + m = m;%a: ⊦ (m + n0) + n1 = m + n0 + n1;%c: ⊦ m + n = n + m;