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Proof step 6 from proof monls-to (previous step)

1 + n0 < n, mon<(⌜nnf⌝ ), nnf(n0 + 1) ≤ nnf(n0) ⊦

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ mon<(nσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1])
Proof step 8:
1 + n0 < n, nnf(n0 + 1) ≤ nnf(n0), ∀ n. ⌜n⌝ + ⌜1⌝ < # ⌜nnf⌝ → ⌜nnf⌝ [n] < ⌜nnf⌝ [n + 1] ⊦

Rule argument

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

Additional information

Used terms: ⌜nnf⌝