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Proof step 115 from proof inv-get-inverse-succ (previous step)

# nσ = ⌜m⌝ + ⌜1⌝ , n1 < m, nσ[n1] ≤ nσ[m], nσ ≠ [] ∧ nσ[0] = 0 ∧ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1), ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ m = 0, nσ[n1 + 1] = nσ[n1] + 1, nσ[n1 + 1] = nσ[n1]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 116:
# nσ = ⌜m⌝ + ⌜1⌝ , nσ[0] = 0, n1 < m, nσ[n1] ≤ nσ[m], ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0, ∀ n. n < m → nσ[n + 1] = nσ[n] ∨ nσ[n + 1] = nσ[n] + 1 ⊦ m = 0, nσ[n1 + 1] = nσ[n1] + 1, nσ[n1 + 1] = nσ[n1], nσ = []

Rule argument

Additional information

Used simplifier rules: less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%eqls-succ: ⊦ n∞0 + ⌜1⌝ = n∞ → (⌜n⌝ + ⌜1⌝ < n∞ ↔ ⌜n⌝ < n∞0); from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf