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Proof step 8 from proof mon-lem (previous step)

σσ[m] + σ = σσ[m1 + m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, mon(σσ), ⌜m1⌝ + ⌜m⌝ < # σσ ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ mon(σσ) ↔ (∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ[n] ⊑ σσ[n + 1])
Proof step 10:
σσ[m] + σ = σσ[m1 + m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → σσ[n] ⊑ σσ[n + 1], ⌜m1⌝ + ⌜m⌝ < # σσ ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Rule argument

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

Additional information

Used terms: σσ