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Proof step 4 from proof mininv-inverts (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) ) ), min(nσ, n + 1) = 1 + min(nσ, n), mon<(nσ) ⊦ nσ[min(nσ, n)] = n

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
min(nσ, n + 1) = 1 + min(nσ, n), mon<(nσ), (# 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) ) ⊦ nσ[min(nσ, n)] = n, nσ = []

Rule argument

Additional information

Used simplifier rules: min-empty: ⊦ min([], n) = 0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat