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Proof step 5 from proof tot (previous step)

∀ m. n0 = m ∨ m < n0 ∨ n0 < m ⊦ n0 +1 = m, m = n0, n0 +1 < m, m < n0

User applied rule all left

And obtained the following new goals

Proof step 6:
∀ m. n0 = m ∨ m < n0 ∨ n0 < m, n0 = m ∨ m < n0 ∨ n0 < m ⊦ n0 +1 = m, m = n0, m < n0, n0 +1 < m

Rule argument

pos: left 1 terms: m abort: F discard: F computed: T

Additional information

Used terms: m