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Proof step 118 from proof inv-get-inverse-between (previous step)

# nσ = ∞, nσ[m0] = nσ[n1] + 1, n1 + m + 1 < ∞.n, nσ[n1] < nσ[m1], stutf(nσ), m0 ≤ m1, nσ[m0] ≤ nσ[m1], ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, (∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[m0]) ∨ (∀ n. nσ[n] ≠ nσ[m0]) ⊦ nσ[n1 + m + 1] = nσ[m0]

User applied rule case distinction

And obtained the following new goals

Proof step 119:
# nσ = ∞, nσ[m0] = nσ[n1] + 1, nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, stutf(nσ), nσ[m0] ≤ nσ[m1], m0 ≤ m1, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ n. ∃ n0. n < n0 ∧ nσ[n0] = nσ[m0] ⊦ nσ[n1 + m + 1] = nσ[m0]
Proof step 122:
# nσ = ∞, nσ[m0] = nσ[n1] + 1, nσ[n1] < nσ[m1], n1 + m + 1 < ∞.n, stutf(nσ), nσ[m0] ≤ nσ[m1], m0 ≤ m1, ∀ m0. m1 < m0 → nσ[m0] = nσ[m1] + 1, ∀ m. nσ[m] = nσ[n1] → ¬ n1 < m, ∀ n. nσ[n] ≠ nσ[m0] ⊦ nσ[n1 + m + 1] = nσ[m0]

Rule argument

pos: left 10

Proof step 118 from proof inv-get-inverse-between (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information