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Proof step 38 from proof inv-ls-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 39:
nσ[n] = n2 + 1, # nσ = ∞, n2 + 1 < nσ[m], n < m, stutf(nσ), ∀ n. nσ[n] ≠ (n2 + 1) + 1, ∀ n0, nσ. n0 < n2 +1 → (∀ n1. nσ[n1] ≠ n0 + 1) ∧ # nσ = ∞ ∧ stutf(nσ) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1