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Proof step 22 from proof min-stutters (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 23:
min(nσ, n) + 1 < min(nσ, n + 1), min(nσ, n) < min(nσ, n + 1), ⌜min(nσ, n + 1)⌝ < # nσ, n < nσ[min(nσ, n + 1)], mon<(nσ), n ≤ nσ[min(nσ, n)], ⌜min(nσ, n)⌝ < # nσ, # nσ = ∞ ∨ ¬ nσ[(# nσ).n - 1] < n, # nσ = ∞ ∨ n < nσ[(# nσ).n - 1], ⌜min(nσ, n) + 1⌝ < # nσ ∧ n < nσ[min(nσ, n) + 1] → min(nσ, n + 1) ≤ min(nσ, n) + 1, ∀ m. ⌜m⌝ < # nσ ∧ n < nσ[m] → min(nσ, n + 1) ≤ m, ∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m ⊦ min(nσ, n + 1) = 1 + min(nσ, n), min(nσ, n + 1) = min(nσ, n), nσ = []

Rule argument

pos: left 10 terms: min(nσ, n) + 1 abort: F discard: F computed: T

Additional information

Used terms: min(nσ, n) + 1