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Proof step 11 from proof mininv-stays-if-not-taken (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 12:
# 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 = []

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [min(nσ, n + 1), m0]

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: min(nσ, n + 1), m0