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Proof step 14 from proof inv-get-inverse (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 15:
∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = n
Proof step 26:
∀ n0. nσ[n0] ≠ n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = n

Rule argument

pos: left 1

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