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Proof step 6 from proof q-add-01 (previous step)

(∃ m1. m0 + m1 = n) ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ ∃ m. (m0 + 1) + m = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 + m = n ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ ∃ m. (m0 + 1) + m = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

Rule argument

Additional information