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Proof step 5 from proof tot (
previous step
)
∀ m. n0 = m ∨ n0 < m ∨ m < n0, ∀ n. (∀ m. n = m ∨ n < m ∨ m < n) → n +1 = m0 ∨ n +1 < m0 ∨ m0 = n ∨ m0 < n ⊦ n0 = m0, n0 < m0, m0 < n0
Heuristic quantifier closing applied rule all left
And closed the goal!
Rule argument
pos: left 1 terms: m0
Additional information
Used terms: m0