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Proof step 17 from proof inv-get-inverse-succ (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 18:
# nσ = ∞, stutf(nσ), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n ⊦
Proof step 29:
# nσ = ∞, stutf(nσ), ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. nσ[n0] ≠ n ⊦

Rule argument

pos: left 5

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