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Proof step 10 from proof monls-ls (previous step)

⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1], ⌜m1⌝ + ⌜m⌝ < # nσ, m1 = 0 ∨ ¬ nσ[m1 + m] ≤ nσ[m] ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 11:
m1 = 0, ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1], ⌜m1⌝ + ⌜m⌝ < # nσ ⊦
Proof step 13:
¬ nσ[m1 + m] ≤ nσ[m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1], ⌜m1⌝ + ⌜m⌝ < # nσ ⊦

Rule argument

pos: left 5

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