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Proof step 20 from proof mininv-inverts-inner-succ (previous step)

mon<(nσ) ∧ n < n + 1 ∧ ⌜n + 1⌝ < # nσ → nσ[n] < nσ[n + 1], ⌜n⌝ + ⌜1⌝ < # nσ, ⌜min(nσ, nσ[n] + 1)⌝ < # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜n⌝ < # nσ, mon<(nσ), # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m, nσ[n] < nσ[n + 1] → ¬ n < min(nσ, nσ[n] + 1) ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
⌜n⌝ + ⌜1⌝ < # nσ, ⌜min(nσ, nσ[n] + 1)⌝ < # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜n⌝ < # nσ, mon<(nσ), nσ[n] < nσ[n + 1], min(nσ, nσ[n] + 1) ≤ n, # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0, nσ[n + 1] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%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%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2