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Proof step 11 from proof mininv-inverts (previous step)

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 = []

User applied rule case distinction

And obtained the following new goals

Proof step 12:
min(nσ, n + 1) = 1 + min(nσ, n), n < nσ[1 + min(nσ, n)], ⌜1⌝ + ⌜min(nσ, n)⌝ < # nσ, ⌜min(nσ, n)⌝ < # nσ, n < nσ[min(nσ, n)], mon<(nσ), ∀ 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 = ∞ ∨ ¬ nσ[(# nσ).n - 1] < n ⊦ nσ[min(nσ, n)] = n, nσ = []

Rule argument

pos: left 5

Additional information

Used simplifier rules: ⊦ m < n → ¬ n < m; (s)