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Proof step 11 from proof monls-ls (previous step)

m1 = 0, ⌜m1⌝ + ⌜1⌝ + ⌜m⌝ < # nσ, nσ[(m1 + 1) + m] ≤ nσ[m], ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1], ⌜m1⌝ + ⌜m⌝ < # nσ ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⌜1⌝ + ⌜m⌝ < # nσ, nσ[1 + m] ≤ nσ[m], ⌜m⌝ < # nσ, ∀ n. ⌜n⌝ + ⌜1⌝ < # nσ → nσ[n] < nσ[n + 1] ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-all-zero: ⊦ n∞ + ⌜0⌝ = n∞; from specification nat-inf