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Proof step 9 from proof mininv-inverts (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
min(nσ, n + 1) = 1 + min(nσ, n), ⌜1⌝ + ⌜min(nσ, n)⌝ < # nσ, n < nσ[1 + min(nσ, n)], mon<(nσ), # nσ = ∞ ∨ n < nσ[(# nσ).n - 1], ∀ m. ⌜m⌝ < # nσ ∧ n < nσ[m] → min(nσ, n) < m, (# 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 = [], nσ[min(nσ, n)] = n, nσ[1 + min(nσ, n)] = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1