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Theorem xorbi1d 1288
Description: Deduction joining an equivalence and a right operand to form equivalence of exclusive-or. (Contributed by Jim Kingdon, 7-Oct-2018.)
Hypothesis
Ref Expression
xorbid.1 (𝜑 → (𝜓𝜒))
Assertion
Ref Expression
xorbi1d (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜃)))

Proof of Theorem xorbi1d
StepHypRef Expression
1 xorbid.1 . . . 4 (𝜑 → (𝜓𝜒))
21orbi1d 715 . . 3 (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜃)))
31anbi1d 446 . . . 4 (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜃)))
43notbid 602 . . 3 (𝜑 → (¬ (𝜓𝜃) ↔ ¬ (𝜒𝜃)))
52, 4anbi12d 450 . 2 (𝜑 → (((𝜓𝜃) ∧ ¬ (𝜓𝜃)) ↔ ((𝜒𝜃) ∧ ¬ (𝜒𝜃))))
6 df-xor 1283 . 2 ((𝜓𝜃) ↔ ((𝜓𝜃) ∧ ¬ (𝜓𝜃)))
7 df-xor 1283 . 2 ((𝜒𝜃) ↔ ((𝜒𝜃) ∧ ¬ (𝜒𝜃)))
85, 6, 73bitr4g 216 1 (𝜑 → ((𝜓𝜃) ↔ (𝜒𝜃)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  wb 102  wo 639  wxo 1282
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-in1 554  ax-in2 555  ax-io 640
This theorem depends on definitions:  df-bi 114  df-xor 1283
This theorem is referenced by:  xorbi12d  1289
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