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Proof step 16 from proof mininv-inverts-inner-succ (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 17:
⌜n⌝ + ⌜1⌝ < # nσ, ⌜min(nσ, nσ[n] + 1)⌝ < # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜n⌝ < # nσ, mon<(nσ), # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m, ⌜n⌝ + ⌜1⌝ < # nσ ∧ nσ[n] < nσ[n + 1] → ¬ n < min(nσ, nσ[n] + 1) ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0
Proof step 23:
⌜min(nσ, nσ[n] + 1)⌝ < # nσ, nσ[n] < nσ[min(nσ, nσ[n] + 1)], ⌜n⌝ < # nσ, mon<(nσ), # nσ = ∞ ∨ nσ[n] < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] < nσ[m] → min(nσ, nσ[n] + 1) ≤ m, ⌜n⌝ + ⌜1⌝ < # nσ ∧ nσ[n] < nσ[n + 1] → ¬ n < min(nσ, nσ[n] + 1) ⊦ ⌜n⌝ + ⌜1⌝ < # nσ, min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0

Rule argument

fma: ⌜n⌝ + ⌜1⌝ < # nσ

Additional information

Used formulas: ⌜n⌝ + ⌜1⌝ < # nσ