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Proof step 80 from proof inv-ls-succ (previous step)

nσ[n0] = n + 1, nσ[m] = n + 1, ⌜m⌝ < # nσ, m < m0, ⌜n0⌝ < # nσ, m < n0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ⌜m0⌝ < # nσ → n < nσ[m0], ∀ m2. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m2 - 1))) → ¬ 2 + n < m2, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1 ⊦ nσ[m0] = n + 1, ⌜m0⌝ < # nσ

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 81:
nσ[m] = n + 1, nσ[n0] = n + 1, n0 < m0, m < n0, ⌜n0⌝ < # nσ, m < m0, ⌜m⌝ < # nσ, m + 1 < m0, stutf(nσ), ∞.n ≤ max∞(λ n0. ⌜n0⌝ < # nσ ∧ nσ[n0] = n).n, ∀ n0. ∃ n1. n0 < n1 ∧ ⌜n1⌝ < # nσ ∧ nσ[n1] = n + 1, ∀ m2. (∃ m1. ⌜m1⌝ < # nσ ∧ nσ[m1] = m2 - 1 ∧ (# nσ ≠ ∞ ∨ ¬(∀ m3. m1 < m3 → nσ[m3] = m2 - 1))) → ¬ 2 + n < m2 ⊦ nσ[m0] = n + 1, ⌜m0⌝ < # nσ

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)%lsle-trans: ⊦ n∞ < n∞1 ∧ ¬ n∞2 < n∞1 → n∞ < n∞2; (forward) from specification nat-inf