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Proof step 57 from proof inv-get-inverse (previous step)

# nσ = ∞, (∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. nσ[n0] ≠ n), n ≤ nσ[m], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 58:
∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, # nσ = ∞, n ≤ nσ[m], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦
Proof step 61:
∀ n0. nσ[n0] ≠ n, # nσ = ∞, n ≤ nσ[m], stutf(nσ), ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

pos: left 2

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