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Proof step 19 from proof eq-leninv (previous step)

m + 1 = nσ[n], ⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ[n]⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n]), nσ[n] = 0

User applied rule case distinction

And obtained the following new goals

Proof step 20:
m + 1 = nσ[n], ⌜nσ[n]⌝ = leninv(nσ), ⌜n⌝ < # nσ, stutf(nσ) ⊦ nσ[n] = 0, # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n])
Proof step 46:
# nσ = ∞, m + 1 = nσ[n], ⌜n⌝ < # nσ, stutf(nσ), ∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n] ⊦ nσ[n] = 0, ⌜nσ[n]⌝ = leninv(nσ)

Rule argument

pos: right 1

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