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Proof step 13 from proof min-stutters (previous step)

# nσ = ⌜min(nσ, n + 1)⌝ , nσ[min(nσ, n + 1) - 1] = 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 elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 14:
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 = []

Rule argument

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

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: min(nσ, n + 1), m0