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Proof step 14 from proof leq-leninv (previous step)

⌜n⌝ < # nσ, n1 + 1 < nσ[n], stutf(nσ), (# nσ = ∞ ⊃ nσ[m] = n1 ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1);nσ[(# nσ).n - 1] = n1) ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 15:
# nσ = ∞, nσ[m] = n1, n1 + 1 < nσ[n], ⌜n⌝ < # nσ, stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦
Proof step 20:
nσ[(# nσ).n - 1] = n1, n1 + 1 < nσ[n], ⌜n⌝ < # nσ, stutf(nσ) ⊦ # nσ = ∞

Rule argument

pos: left 4

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