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Proof step 3 from proof mon-eq (previous step)

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

User applied rule insert lemma

And obtained the following new goals

Proof step 4:
⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ[m] < nσ[n]
Proof step 5:
nσ[n] = nσ[n0], ⌜n⌝ < # nσ, ⌜n0⌝ < # nσ, mon<(nσ), mon<(nσ) ∧ n < n0 ∧ ⌜n0⌝ < # nσ → nσ[n] < nσ[n0] ⊦ n = n0

Rule argument

spec: <toplevel> name: monls-ls seq: ⊦ mon<(nσ) ∧ m < n ∧ ⌜n⌝ < # nσ → nσ[m] < nσ[n] subst: [nσ, m, n] → [nσ, n, n0]

Additional information

Used terms: nσ, n, n0