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Proof step 8 from proof maxinf-skolemf (
previous step
)
np(n0) ⊦ ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), ∃ n0. n < n0 ∧ np(n0), ∃ n∞. (n∞ = ∞ ⊃ (∀ n. ∃ n0. n < n0 ∧ np(n0)) ∨ (∀ n. ¬ np(n));np(n∞.n) ∧ (∀ m. np(m) → ¬ n∞ < ⌜m⌝ ))
User applied rule weakening
And obtained the following new goals
Proof step 9
:
np(n0) ⊦ ∃ n. np(n) ∧ (∀ n0. n < n0 → ¬ np(n0)), ∃ n0. n < n0 ∧ np(n0)
Rule argument
poslist: right 3
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