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Proof step 11 from proof min-skolemf (previous step)

mon<(nσ), ∀ n0, nσ. n0 < 0 → mon<(nσ) → nσ = [] ∨ (∃ m0. ⌜m0⌝ < # nσ ∧ n0 ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n0 ≤ nσ[m] → m0 ≤ m)) ∨ # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n0 ⊦ nσ = [], ∃ m0. ⌜m0⌝ < # nσ ∧ (∀ m. ⌜m⌝ < # nσ → m0 ≤ m)

User applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: 0 abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ ⌜0⌝ < # nσ ↔ nσ ≠ []; (s)% ⊦ 0 ≤ m; (s) Used terms: 0