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Proof step 11 from proof ls-ext (previous step)

∀ m. nx0 < m ↔ (∀ n. n ∈ nx0 → n < m), ∀ n. n = m0 ∨ n ∈ nx0 → n < m ⊦ m0 ' + nx0 < m

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 12:
m0 = m0 ∨ m0 ∈ nx0 → m0 < m, ∀ n. n ∈ nx0 → n < m, ∀ m. nx0 < m ↔ (∀ n. n ∈ nx0 → n < m) ⊦ m0 ' + nx0 < m

Rule argument

pos: left 2 terms: m0

Additional information

Used terms: m0