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Proof step 15 from proof mininv-stays-if-not-taken (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 16:
# nσ = ∞ ∨ ¬ nσ[(# nσ).n - 1] < n, ⌜min(nσ, n)⌝ < # nσ ∧ n ≤ nσ[min(nσ, n)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m), ⌜min(nσ, n + 1)⌝ < # nσ, n < nσ[min(nσ, n + 1)], mon<(nσ), # nσ = ∞ ∨ n < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ n < nσ[m] → min(nσ, n + 1) ≤ m, ∀ m. ⌜m⌝ < # nσ → nσ[m] ≠ n ⊦ min(nσ, n + 1) = min(nσ, n), nσ = [], nσ[min(nσ, n + 1)] = 0

Rule argument

pos: left 7

Additional information

Used simplifier rules: ls-02: ⊦ m < n → ¬ n < m; from specification nat-basic1