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Proof step 5 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 quantifier closing applied rule all left

And obtained the following new goals

Proof step 6:
m + m1 + 1 < n, n + n0 < m, m + m1 < n ∧ n + n0 < m → m = 0 ∨ n = 0, ∀ m, n, n0. m + m1 < n ∧ n + n0 < m → m = 0 ∨ n = 0 ⊦ m = 0, n = 0

Rule argument

pos: left 3 terms: m, n, n0

Additional information

Used terms: m, n, n0