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Proof step 4 from proof l-05 (previous step)

n + n0 < m, m + m1 +1 < n, ∀ m, n, n0. m + m1 < n ∧ n + n0 < m → m = 0 ∨ n = 0 ⊦ m = 0, n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n + n0 < m, m + m1 + 1 < n, ∀ m, n, n0. m + m1 < n ∧ n + n0 < m → m = 0 ∨ n = 0 ⊦ m = 0, n = 0

Rule argument

Additional information

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