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Proof step 26 from proof inv-ls-succ (previous step)

# nσ = ∞, stutf(nσ), ∀ n0. nσ[n0] ≠ n + 1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0, nσ. n0 < n → (∀ n1. nσ[n1] ≠ n0 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

User applied rule constructor cut

And obtained the following new goals

Proof step 27:
n = 0, # nσ = ∞, stutf(nσ), ∀ n0. nσ[n0] ≠ n + 1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0, nσ. n0 < n → (∀ n1. nσ[n1] ≠ n0 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦
Proof step 34:
n = n2 +1, # nσ = ∞, stutf(nσ), ∀ n0. nσ[n0] ≠ n + 1, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0, nσ. n0 < n → (∀ n1. nσ[n1] ≠ n0 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

Rule argument

var: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;