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Proof step 9 from proof mininv-inverts (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 10:
min(nσ, n + 1) = 1 + min(nσ, n), ⌜1⌝ + ⌜min(nσ, n)⌝ < # nσ, n < nσ[1 + min(nσ, n)], mon<(nσ), (# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ min(nσ, n) = (# nσ).n ;⌜min(nσ, n)⌝ < # nσ ∧ n < nσ[min(nσ, n)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m) ), ∀ m. ⌜m⌝ < # nσ ∧ n < nσ[m] → min(nσ, n) < m, # nσ = ∞ ∨ n < nσ[(# nσ).n - 1] ⊦ nσ[min(nσ, n)] = n, nσ = []

Rule argument

pos: left 4

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m = m + n ↔ n = 0; (s)