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Proof step 2 from proof eq-leninv (previous step)

⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ[n]⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n])

User applied rule cut formula

And obtained the following new goals

Proof step 3:
nσ[n] = 0, ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ[n]⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n])
Proof step 15:
⌜n⌝ < # nσ, stutf(nσ) ⊦ nσ[n] = 0, ⌜nσ[n]⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n])

Rule argument

fma: nσ[n] = 0

Additional information

Used formulas: nσ[n] = 0