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Proof step 12 from proof mininv-stays-if-not-taken (previous step)

min(nσ, n + 1) = m0 + 1, nσ[m0] = n, # nσ = ⌜min(nσ, n + 1)⌝ , n ≤ nσ[min(nσ, n)], mon<(nσ), min(nσ, n) < min(nσ, n + 1), ∀ m. m < min(nσ, nσ[m] + 1) → nσ[m] ≠ n, ∀ m. m < min(nσ, n + 1) ∧ n ≤ nσ[m] → min(nσ, n) ≤ m ⊦ nσ = [], min(nσ, n + 1) = min(nσ, n), min(nσ, n + 1) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# nσ = ⌜m0⌝ + ⌜1⌝ , min(nσ, nσ[m0] + 1) = m0 + 1, min(nσ, nσ[m0]) ≤ m0, nσ[m0] ≤ nσ[min(nσ, nσ[m0])], mon<(nσ), ∀ m. ¬ m0 < m ∧ nσ[m0] ≤ nσ[m] → min(nσ, nσ[m0]) ≤ m, ∀ m. m < min(nσ, nσ[m] + 1) → nσ[m] ≠ nσ[m0] ⊦ nσ = [], m0 + 1 = min(nσ, nσ[m0])

Rule argument

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1