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Proof step 14 from proof mininv-inverts-inner (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 15:
⌜min(nσ, nσ[n])⌝ < # nσ, nσ[n] ≤ nσ[min(nσ, nσ[n])], ⌜n⌝ < # nσ, mon<(nσ), # nσ = ∞ ∨ ¬ nσ[(# nσ).n - 1] < nσ[n], ⌜n⌝ < # nσ ∧ nσ[n] ≤ nσ[n] → min(nσ, nσ[n]) ≤ n, ∀ m. ⌜m⌝ < # nσ ∧ nσ[n] ≤ nσ[m] → min(nσ, nσ[n]) ≤ m ⊦ min(nσ, nσ[n]) = n, nσ = []

Rule argument

pos: left 6 terms: n abort: F discard: F computed: T

Additional information

Used terms: n