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

∀ nσ. mon<(nσ) ∧ nσ[0] = 0 ∧ ⌜n0 + 1⌝ < # nσ → (¬ n0 < nσ[n0 + 1] ↔ false) ⊦ mon<(nσ) ∧ nσ[0] = 0 ∧ ⌜n0 +1 + 1⌝ < # nσ → (¬ n0 +1 < nσ[n0 +1 + 1] ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
nσ[0] = 0, mon<(nσ), ⌜2⌝ + ⌜n0⌝ < # nσ, nσ[2 + n0] ≤ n0 + 1, ∀ nσ0. mon<(nσ0) ∧ nσ0[0] = 0 ∧ ⌜n0⌝ + ⌜1⌝ < # nσ0 → n0 < nσ0[n0 + 1] ⊦

Rule argument

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2