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Proof step 5 from proof monls-to (previous step)

mon<(⌜nnf⌝ ), 1 + n0 < n, nnf(n0 + 1) ≤ nnf(n0) ⊦ n = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ mon<(nσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1])
Proof step 7:
∀ n. ⌜n⌝ + ⌜1⌝ < # ⌜nnf⌝ → ⌜nnf⌝ [n] < ⌜nnf⌝ [n + 1], 1 + n0 < n, nnf(n0 + 1) ≤ nnf(n0) ⊦ n = 0

Rule argument

spec: inst: name: monls-def seq: ⊦ mon<(nσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1]) subst: [nσ] → [⌜nnf⌝ ] flags: rewrite, current

Additional information

Used terms: ⌜nnf⌝