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

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
m1 + 1 = n0, n = n0 + m, ⌜n⌝ < # σσ, mon(σσ), m ≤ n, ∀ n, m, σσ. m1 = n - m ∧ ⌜n⌝ < # σσ ∧ mon(σσ) ∧ m ≤ n → (∃ σ1. σσ[m] + σ1 = σσ[n]) ⊦ ∃ σ1. σσ[m] + σ1 = σσ[n]

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, m, n1]

Additional information

Used simplifier rules: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n1