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Proof step 2 from proof mininv-inverts-inner-succ (previous step)

mon<(nσ), ⌜n⌝ < # nσ ⊦ min(nσ, nσ[n] + 1) = n + 1

User applied rule insert lemma

And obtained the following new goals

Proof step 3:
⊦ mon<(nσ) → (nσ = [] ⊃ min(nσ, n) = 0 ;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ min(nσ, n) = (# nσ).n ;⌜min(nσ, n)⌝ < # nσ ∧ n ≤ nσ[min(nσ, n)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m) ) )
Proof step 4:
mon<(nσ) → (nσ = [] ⊃ min(nσ, nσ[n] + 1) = 0 ;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < nσ[n] + 1 ⊃ min(nσ, nσ[n] + 1) = (# nσ).n ;⌜min(nσ, nσ[n] + 1)⌝ < # nσ ∧ nσ[n] + 1 ≤ nσ[min(nσ, nσ[n] + 1)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ nσ[n] + 1 ≤ nσ[m] → min(nσ, nσ[n] + 1) ≤ m) ) ), mon<(nσ), ⌜n⌝ < # nσ ⊦ min(nσ, nσ[n] + 1) = n + 1

Rule argument

spec: inst: name: min-def seq: ⊦ mon<(nσ) → (nσ = [] ⊃ min(nσ, n) = 0 ;(# nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n ⊃ min(nσ, n) = (# nσ).n ; ⌜min(nσ, n)⌝ < # nσ ∧ n ≤ nσ[min(nσ, n)] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → min(nσ, n) ≤ m) ) ) subst: [nσ, n] → [nσ, nσ[n] + 1] flags: current, []

Additional information

Used terms: nσ, nσ[n] + 1