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Proof step 196 from proof inv-get-inverse-between (previous step)

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

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: discard

Additional information

Used simplifier rules: add-ls-cancel-num-04: ⊦ m ≤ n → (⌜(* m)⌝ + n∞2 < ⌜(* n)⌝ + n∞1 ↔ n∞2 < ⌜n - m⌝ + n∞1); from specification nat-inf%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%ls-natinf-to-nat-2-1: ⊦ ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n + m < n0; from specification nat-inf%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat