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Theorem bianass 655
Description: An inference to merge two lists of conjuncts. (Contributed by Giovanni Mascellani, 23-May-2019.)
Hypothesis
Ref Expression
bianass.1 (𝜑 ↔ (𝜓 ∧ 𝜒))
Assertion
Ref Expression
bianass ((𝜂 ∧ 𝜑) ↔ ((𝜂 ∧ 𝜓) ∧ 𝜒))

Proof of Theorem bianass
StepHypRef Expression
1 bianass.1 . . 3 (𝜑 ↔ (𝜓 ∧ 𝜒))
21anbi2i 635 . 2 ((𝜂 ∧ 𝜑) ↔ (𝜂 ∧ (𝜓 ∧ 𝜒)))
3 anass 474 . 2 (((𝜂 ∧ 𝜓) ∧ 𝜒) ↔ (𝜂 ∧ (𝜓 ∧ 𝜒)))
42, 3bitr4i 281 1 ((𝜂 ∧ 𝜑) ↔ ((𝜂 ∧ 𝜓) ∧ 𝜒))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This proof depends on definitions:  df-bi 210  df-an 402
This theorem is used by:  bianassc  656  an12  658  an4  669  cnvresima  6224  elcncf1di  25196  nb3grpr2  29946  dfpth2  30296  wwlksnextwrd  30468  cusgr3cyclex  35880  satfvsuclem2  36094  bj-prmoore  38004  bj-imdirco  38079  redundpim3  39614  isthincd2  50489
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