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Proof step 6 from proof min-skolemf (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
mon<(nσ) ⊦ nσ = [], ∃ m0. ⌜m0⌝ < # nσ ∧ n ≤ nσ[m0] ∧ (∀ m. ⌜m⌝ < # nσ ∧ n ≤ nσ[m] → m0 ≤ m), # nσ ≠ ∞ ∧ nσ[(# nσ).n - 1] < n

Proof step 6 from proof min-skolemf (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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