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Proof step 5 from proof mininv-inverts-inner-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⌜n⌝ < # nσ, mon<(nσ), (# nσ ≠ ∞ ∧ ¬ nσ[n] < nσ[(# nσ).n - 1] ⊃ min(nσ, nσ[n] + 1) = (# nσ).n ;⌜min(nσ, nσ[n] + 1)⌝ < # nσ ∧ nσ[n] < nσ[min(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 = []

Rule argument

Additional information

Used simplifier rules: min-empty: ⊦ min([], n) = 0;% ⊦ ¬ m < 0; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ # [] = ⌜0⌝ ; (s)% ⊦ m + 1 ≠ 0; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)