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Proof step 6 from proof monls-ls (previous step)

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] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, mon<(nσ), nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n0, m0, nσ0. m1 ≠ n0 - m0 ∨ ¬ m0 < n0 ∨ ¬ ⌜n0⌝ < # nσ0 ∨ ¬ mon<(nσ0) ∨ ¬ nσ0[n0] ≤ nσ0[m0] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)% ⊦ ¬ m + n < m; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m < n → m ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic