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Proof step 3 from proof mininv-stays-if-not-taken (previous step)

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) ) ), mon<(nσ), ∀ m. ⌜m⌝ < # nσ → nσ[m] ≠ n ⊦ min(nσ, n + 1) = min(nσ, n)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
mon<(nσ), ∀ m. ⌜m⌝ < # nσ → nσ[m] ≠ n, (# 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) ) ⊦ min(nσ, n + 1) = min(nσ, n), nσ = []

Rule argument

Additional information

Used simplifier rules: min-empty: ⊦ min([], n) = 0;