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Proof step 14 from proof min-stutters (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# nσ = ⌜m0⌝ + ⌜1⌝ , min(nσ, nσ[m0] + 1) = m0 + 1, mon<(nσ), nσ[m0] ≤ nσ[min(nσ, nσ[m0])], min(nσ, nσ[m0]) < m0, ∀ m. m < m0 + 1 ∧ nσ[m0] ≤ nσ[m] → min(nσ, nσ[m0]) ≤ m ⊦ m0 = min(nσ, nσ[m0]), m0 + 1 = min(nσ, nσ[m0]), nσ = []

Rule argument

Additional information

Used simplifier rules: neq-01: ⊦ m ≠ n → (m < n + 1 ↔ m < n); from specification nat-basic2%l-06: ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%add-fin: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; from specification nat-inf%succ-03: ⊦ m < n ∧ m + 1 ≠ n → m + 1 < n; (forward) from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1