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Proof step 11 from proof mininv-inverts-inner (previous step)

# nσ = ⌜m⌝ + ⌜1⌝ , min(nσ, nσ[n]) = m + 1, n ≤ m, nσ[m] < nσ[n], mon<(nσ) ⊦ nσ[n] = 0, nσ = [], m + 1 = n

User applied rule insert lemma

And obtained the following new goals

Rule argument

spec: inst: name: monls-ls seq: ⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ[m] < nσ[n] subst: [nσ, m, n] → [nσ, n, m] flags: current, []

Additional information

Used simplifier rules: 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%inf-inseq-reduce: ⊦ n∞ = ⌜m⌝ + ⌜n0⌝ → (⌜n⌝ < n∞ ↔ n < m + n0); from specification nat-inf%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2 Used terms: nσ, n, m