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

m1 + 1 = n - m, m < n, ⌜n⌝ < # nσ, mon<(nσ), nσ[n] ≤ nσ[m], ∀ n0, m0, nσ0. m1 ≠ n0 - m0 ∨ ¬ m0 < n0 ∨ ¬ ⌜n0⌝ < # nσ0 ∨ ¬ mon<(nσ0) ∨ ¬ nσ0[n0] ≤ nσ0[m0] ⊦

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
m1 + 1 = n1, n = n1 + m, ⌜n⌝ < # nσ, m < n, mon<(nσ), nσ[n] ≤ nσ[m], m ≤ n, ∀ n0, m0, nσ0. m1 ≠ n0 - m0 ∨ ¬ m0 < n0 ∨ ¬ ⌜n0⌝ < # nσ0 ∨ ¬ mon<(nσ0) ∨ ¬ nσ0[n0] ≤ nσ0[m0] ⊦

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: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < n → ¬ n < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, m, n1