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Proof step 5 from proof mon-lem (previous step)

n = n0 + m, m1 + 1 = n0, mon(σσ), m ≤ n, ⌜n⌝ < # σσ, ∀ n, m, σσ. m1 = n - m ∧ mon(σσ) ∧ m ≤ n ∧ ⌜n⌝ < # σσ → (∃ σ1. σσ[m] + σ1 = σσ[n]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
mon(σσ), ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # σσ, ∀ n, m, σσ. m1 = n - m ∧ mon(σσ) ∧ m ≤ n ∧ ⌜n⌝ < # σσ → (∃ σ1. σσ[m] + σ1 = σσ[n]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[(m1 + 1) + m]

Rule argument

Additional information

Used simplifier rules: add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat