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Proof step 5 from proof monle-leq (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

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

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