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Proof step 39 from proof min-skolemf (previous step)

⌜m0⌝ + ⌜1⌝ = # nσ, nσ[m0] < nσ[(# nσ).n - 1], mon<(nσ), ∀ m. ¬ m0 < m ∧ nσ[m0] ≤ nσ[m] → m0 ≤ m, ∀ n0, nσ0. n0 < nσ[m0] +1 → mon<(nσ0) → nσ0 = [] ∨ (∃ m0. ⌜m0⌝ < # nσ0 ∧ n0 ≤ nσ0[m0] ∧ (∀ m. ⌜m⌝ < # nσ0 ∧ n0 ≤ nσ0[m] → m0 ≤ m)) ∨ # nσ0 ≠ ∞ ∧ nσ0[(# nσ0).n - 1] < n0 ⊦ nσ = [], ∃ m. ¬ m0 < m ∧ nσ[m0] < nσ[m] ∧ (∀ m1. ¬ m0 < m1 ∧ nσ[m0] < nσ[m1] → m ≤ m1)

User applied rule insert equation

And closed the goal!

Rule argument

pos: left 1 flags: rotate, discard

Additional information

Used simplifier rules: ⊦ ⌜m⌝ ≠ ∞; (s)%n∞ ≠ ∞ ⊦ (⌜m⌝ + n∞).n = m + n∞.n; (s)% ⊦ ⌜m⌝ .n = m; (s)%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ ¬ n < n; (s)