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Theorem xorbi2d 1314
Description: Deduction joining an equivalence and a left operand to form equivalence of exclusive-or. (Contributed by Jim Kingdon, 7-Oct-2018.)
Hypothesis
Ref Expression
xorbid.1 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
xorbi2d (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))

Proof of Theorem xorbi2d
StepHypRef Expression
1 xorbid.1 . . . 4 (𝜑 → (𝜓𝜒))
21orbi2d 737 . . 3 (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
31anbi2d 452 . . . 4 (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
43notbid 625 . . 3 (𝜑 → (¬ (𝜃𝜓) ↔ ¬ (𝜃𝜒)))
52, 4anbi12d 457 . 2 (𝜑 → (((𝜃𝜓) ∧ ¬ (𝜃𝜓)) ↔ ((𝜃𝜒) ∧ ¬ (𝜃𝜒))))
6 df-xor 1310 . 2 ((𝜃𝜓) ↔ ((𝜃𝜓) ∧ ¬ (𝜃𝜓)))
7 df-xor 1310 . 2 ((𝜃𝜒) ↔ ((𝜃𝜒) ∧ ¬ (𝜃𝜒)))
85, 6, 73bitr4g 221 1 (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 102  wb 103  wo 662  wxo 1309
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 577  ax-in2 578  ax-io 663
This theorem depends on definitions:  df-bi 115  df-xor 1310
This theorem is referenced by:  xorbi12d  1316
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