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

np(n0), np(n1), ∀ n0. n1 < n0 → ¬ np(n0) ⊦ (⌜n1⌝ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));np(⌜n1⌝ .n) ∧ (∀ m. np(m) → ¬ ⌜n1⌝ < ⌜m⌝ )), ∃ n0. n < n0 ∧ np(n0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n1 < m, np(m), np(n0), np(n1), ∀ n0. n1 < n0 → ¬ np(n0) ⊦ ∃ n0. n < n0 ∧ np(n0)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ ⌜m⌝ .n = m; (s)% ⊦ ⌜m⌝ ≠ ∞; (s)