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Theorem hbimtg 36538
Description: A more general and closed form of hbim 2333. (Contributed by Scott Fenton, 13-Dec-2010.)
Assertion
Ref Expression
hbimtg ((∀𝑥(𝜑 → ∀𝑥𝜒) ∧ (𝜓 → ∀𝑥𝜃)) → ((𝜒 → 𝜓) → ∀𝑥(𝜑 → 𝜃)))

Proof of Theorem hbimtg
StepHypRef Expression
1 hbntg 36537 . . . 4 (∀𝑥(𝜑 → ∀𝑥𝜒) → (¬ 𝜒 → ∀𝑥 ¬ 𝜑))
2 pm2.21 124 . . . . 5 (¬ 𝜑 → (𝜑 → 𝜃))
32alimi 1844 . . . 4 (∀𝑥 ¬ 𝜑 → ∀𝑥(𝜑 → 𝜃))
41, 3syl6 36 . . 3 (∀𝑥(𝜑 → ∀𝑥𝜒) → (¬ 𝜒 → ∀𝑥(𝜑 → 𝜃)))
54adantr 486 . 2 ((∀𝑥(𝜑 → ∀𝑥𝜒) ∧ (𝜓 → ∀𝑥𝜃)) → (¬ 𝜒 → ∀𝑥(𝜑 → 𝜃)))
6 ala1 1846 . . . 4 (∀𝑥𝜃 → ∀𝑥(𝜑 → 𝜃))
76imim2i 17 . . 3 ((𝜓 → ∀𝑥𝜃) → (𝜓 → ∀𝑥(𝜑 → 𝜃)))
87adantl 487 . 2 ((∀𝑥(𝜑 → ∀𝑥𝜒) ∧ (𝜓 → ∀𝑥𝜃)) → (𝜓 → ∀𝑥(𝜑 → 𝜃)))
95, 8jad 189 1 ((∀𝑥(𝜑 → ∀𝑥𝜒) ∧ (𝜓 → ∀𝑥𝜃)) → ((𝜒 → 𝜓) → ∀𝑥(𝜑 → 𝜃)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 401  ∀wal 1568
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2178  ax-12 2213
This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813
This theorem is used by:  hbimg  36541
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