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Theorem xorbi2d 1255
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 691 . . 3 (φ → ((θ ψ) ↔ (θ χ)))
31anbi2d 440 . . . 4 (φ → ((θ ψ) ↔ (θ χ)))
43notbid 579 . . 3 (φ → (¬ (θ ψ) ↔ ¬ (θ χ)))
52, 4anbi12d 445 . 2 (φ → (((θ ψ) ¬ (θ ψ)) ↔ ((θ χ) ¬ (θ χ))))
6 df-xor 1252 . 2 ((θψ) ↔ ((θ ψ) ¬ (θ ψ)))
7 df-xor 1252 . 2 ((θχ) ↔ ((θ χ) ¬ (θ χ)))
85, 6, 73bitr4g 212 1 (φ → ((θψ) ↔ (θχ)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4   wa 97  wb 98   wo 616  wxo 1251
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 532  ax-in2 533  ax-io 617
This theorem depends on definitions:  df-bi 110  df-xor 1252
This theorem is referenced by:  xorbi12d  1257
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