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Proof step 19 from proof mininv-stays-if-not-taken (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 20:
⌜min(nσ, n)⌝ < # nσ, n ≤ nσ[min(nσ, n)], ⌜min(nσ, n + 1)⌝ < # nσ, n < nσ[min(nσ, n + 1)], mon<(nσ), min(nσ, n) < min(nσ, n + 1), ∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m, # nσ = ∞ ∨ n < nσ[(# nσ).n - 1], ⌜min(nσ, n)⌝ < # nσ ∧ n < nσ[min(nσ, n)] → min(nσ, n + 1) ≤ min(nσ, n), ∀ 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 + 1) = min(nσ, n), nσ = [], nσ[min(nσ, n + 1)] = 0, min(nσ, n + 1) = 0

Rule argument

pos: left 9 terms: min(nσ, n) abort: F discard: F computed: T

Additional information

Used terms: min(nσ, n)