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Proof step 13 from proof mininv-stays-if-not-taken (previous step)

# 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])

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 7 terms: m0

Additional information

Used simplifier rules: lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%eq: ⊦ m = n + 1 → ¬ m ≤ n; from specification nat Used terms: m0