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Proof step 9 from proof monls-ls (previous step)

⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, ⌜m1⌝ + ⌜m⌝ < # nσ, mon<(nσ), nσ[(m1 + 1) + m] ≤ nσ[m], ¬ nσ[m1 + m] ≤ nσ[m] ∨ m1 = 0 ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ mon<(nσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1])
Proof step 11:
⌜m1⌝ + ⌜m⌝ < # nσ, ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1], ¬ nσ[m1 + m] ≤ nσ[m] ∨ m1 = 0 ⊦

Rule argument

spec: <toplevel> name: monls-def seq: ⊦ mon<(nσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1]) subst: [nσ] → [nσ] flags: <all paths>

Additional information

Used terms: nσ