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Proof step 11 from proof inv-get-inverse (previous step)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[max∞(λ n0. nσ[n0] = n).n] = n

User applied rule cut formula

And obtained the following new goals

Proof step 12:
# nσ = ∞, max∞(λ n0. nσ[n0] = n) = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[max∞(λ n0. nσ[n0] = n).n] = n
Proof step 39:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[max∞(λ n0. nσ[n0] = n).n] = n, max∞(λ n0. nσ[n0] = n) = ∞

Rule argument

fma: max∞(λ n0. nσ[n0] = n) = ∞

Additional information

Used formulas: max∞(λ n0. nσ[n0] = n) = ∞