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Proof step 21 from proof inv-inverts (previous step)

nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), m < n → nσ[n] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

User applied rule cut formula

And obtained the following new goals

Proof step 22:
m < n, nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), m < n → nσ[n] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n
Proof step 24:
nσ[n + 1] = nσ[n] + 1, # nσ = ∞, max∞(λ n0. nσ[n0] = nσ[n]) = ⌜n⌝ , stutf(nσ), m < n → nσ[n] = nσ[m] + 1, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦ m < n, ((λ m0. max∞(λ n0. nσ[n0] = m0).n) to nσ[m] + 1)[nσ[n]] = n

Rule argument

fma: m < n

Additional information

Used formulas: m < n