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Proof step 10 from proof natset-ls-add-max-ls (previous step)

(m + 1 < n ⊃ n;m + 1) ≤ n0, ∀ n0. n0 ∈ ns → n0 < n, n0 = m ∨ n0 ∈ ns ⊦

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 11:
m + 1 < n, n ≤ n0, ∀ n0. n0 ∈ ns → n0 < n, n0 = m ∨ n0 ∈ ns ⊦
Proof step 13:
m + 1 ≤ n0, ∀ n0. n0 ∈ ns → n0 < n, n0 = m ∨ n0 ∈ ns ⊦ m + 1 < n

Rule argument

pos: left 1

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