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Proof step 35 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. nσ[n] ≠ (n2 + 1) + 1, ∀ n0, nσ. n0 < n2 +1 → # nσ = ∞ ∧ stutf(nσ) ∧ (∀ n1. nσ[n1] ≠ n0 + 1) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)