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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ n → (n < m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m + 1 ≠ 0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)