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Proof step 5 from proof mon-eq (previous step)

nσ[n] = nσ[n0], ⌜n0⌝ < # nσ, ⌜n⌝ < # nσ, mon<(nσ), n0 < n ⊦ n = n0, n = 0

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, n0, n] flags: current, []

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1 Used terms: nσ, n0, n