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Proof step 23 from proof leninv-inf (previous step)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ ∃ n1. n0 < n1 ∧ (∃ m. nσ[m] = n1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n1 - 1))

User applied rule cut formula

And obtained the following new goals

Proof step 24:
∀ n. ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ ∃ n1. n0 < n1 ∧ (∃ m. nσ[m] = n1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n1 - 1))
Proof step 36:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ ∀ n. ∃ m. nσ[m] = n, ∃ n1. n0 < n1 ∧ (∃ m. nσ[m] = n1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n1 - 1))

Rule argument

fma: ∀ n. ∃ m. nσ[m] = n

Additional information

Used formulas: ∀ n. ∃ m. nσ[m] = n