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Proof step 10 from proof mon-lem (previous step)

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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
σσ[m] + σ = σσ[m1 + m], ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ⌜m1⌝ + ⌜m⌝ < # σσ, ∀ n. ⌜n⌝ + ⌜1⌝ < # σσ → (∃ σ1. σσ[n] + σ1 = σσ[n + 1]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Rule argument

Additional information

Used simplifier rules: prefix-def: ⊦ σ ⊑ σ0 ↔ (∃ σ1. σ + σ1 = σ0);