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Proof step 15 from proof mininv-inverts-inner-succ (previous step)

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

User applied rule all left

And obtained the following new goals

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

Rule argument

pos: left 5 terms: n + 1 abort: F discard: F computed: F

Additional information

Used terms: n + 1