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Proof step 4 from proof q-add-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ n. (∃ m1. m0 + m1 = n) ∨ n ≠ m0 ∧ ¬ m0 < n ⊦ ∃ m. (m0 + 1) + m = n, n ≠ m0 + 1 ∧ ¬ m0 + 1 < n

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1;%a: ⊦ (m + n0) + n1 = m + n0 + n1;