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Proof step 14 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 ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0

User applied rule all left

And obtained the following new goals

Proof step 15:
⌜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], ⌜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 ⊦ min(nσ, nσ[n] + 1) = n + 1, nσ = [], nσ[min(nσ, nσ[n] + 1)] = 0

Rule argument

pos: left 6 terms: n + 1 abort: F discard: F computed: F

Additional information

Used terms: n + 1