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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
np(n1), np(n0), n1 < m, np(m), ∀ n0. n1 < n0 → ¬ np(n0) ⊦ m = 0, ∃ n0. n < n0 ∧ np(n0)

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%less-fin: ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; from specification nat-inf%sel: ⊦ ⌜m⌝ .n = m; from specification nat-inf-basic%disj: ⊦ ⌜m⌝ ≠ ∞; from specification nat-inf-basic