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Proof step 8 from proof leninv-inf (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 9:
# nσ = ∞, stutf(nσ) ⊦ (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])
Proof step 49:
stutf(nσ) ⊦ # nσ = ∞, (∀ n. ∃ n0. n < n0 ∧ (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n0 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n0 - 1)))) ∨ (∀ n. ¬(n ≠ 0 → (∃ m. ⌜m⌝ < # nσ ∧ nσ[m] = n - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m0. m < m0 → nσ[m0] = n - 1))))) ↔ # nσ = ∞ ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0])

Rule argument

fma: # nσ = ∞

Additional information

Used formulas: # nσ = ∞