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Theorem bj-nnford 34235
 Description: Nonfreeness in both disjuncts implies nonfreeness in the disjunction, deduction form. See comments for bj-nnfor 34234 and bj-nnfand 34233. (Contributed by BJ, 2-Dec-2023.) (Proof modification is discouraged.)
Hypotheses
Ref Expression
bj-nnford.1 (𝜑 → Ⅎ'𝑥𝜓)
bj-nnford.2 (𝜑 → Ⅎ'𝑥𝜒)
Assertion
Ref Expression
bj-nnford (𝜑 → Ⅎ'𝑥(𝜓𝜒))

Proof of Theorem bj-nnford
StepHypRef Expression
1 19.43 1883 . . 3 (∃𝑥(𝜓𝜒) ↔ (∃𝑥𝜓 ∨ ∃𝑥𝜒))
2 bj-nnford.1 . . . . 5 (𝜑 → Ⅎ'𝑥𝜓)
32bj-nnfed 34218 . . . 4 (𝜑 → (∃𝑥𝜓𝜓))
4 bj-nnford.2 . . . . 5 (𝜑 → Ⅎ'𝑥𝜒)
54bj-nnfed 34218 . . . 4 (𝜑 → (∃𝑥𝜒𝜒))
63, 5orim12d 962 . . 3 (𝜑 → ((∃𝑥𝜓 ∨ ∃𝑥𝜒) → (𝜓𝜒)))
71, 6syl5bi 245 . 2 (𝜑 → (∃𝑥(𝜓𝜒) → (𝜓𝜒)))
82bj-nnfad 34216 . . . 4 (𝜑 → (𝜓 → ∀𝑥𝜓))
94bj-nnfad 34216 . . . 4 (𝜑 → (𝜒 → ∀𝑥𝜒))
108, 9orim12d 962 . . 3 (𝜑 → ((𝜓𝜒) → (∀𝑥𝜓 ∨ ∀𝑥𝜒)))
11 19.33 1885 . . 3 ((∀𝑥𝜓 ∨ ∀𝑥𝜒) → ∀𝑥(𝜓𝜒))
1210, 11syl6 35 . 2 (𝜑 → ((𝜓𝜒) → ∀𝑥(𝜓𝜒)))
13 df-bj-nnf 34211 . 2 (Ⅎ'𝑥(𝜓𝜒) ↔ ((∃𝑥(𝜓𝜒) → (𝜓𝜒)) ∧ ((𝜓𝜒) → ∀𝑥(𝜓𝜒))))
147, 12, 13sylanbrc 586 1 (𝜑 → Ⅎ'𝑥(𝜓𝜒))
 Colors of variables: wff setvar class Syntax hints:   → wi 4   ∨ wo 844  ∀wal 1536  ∃wex 1781  Ⅎ'wnnf 34210 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811 This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-bj-nnf 34211 This theorem is referenced by: (None)
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