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Theorem xorbi2d 1358
 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 779 . . 3 (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
31anbi2d 459 . . . 4 (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
43notbid 656 . . 3 (𝜑 → (¬ (𝜃𝜓) ↔ ¬ (𝜃𝜒)))
52, 4anbi12d 464 . 2 (𝜑 → (((𝜃𝜓) ∧ ¬ (𝜃𝜓)) ↔ ((𝜃𝜒) ∧ ¬ (𝜃𝜒))))
6 df-xor 1354 . 2 ((𝜃𝜓) ↔ ((𝜃𝜓) ∧ ¬ (𝜃𝜓)))
7 df-xor 1354 . 2 ((𝜃𝜒) ↔ ((𝜃𝜒) ∧ ¬ (𝜃𝜒)))
85, 6, 73bitr4g 222 1 (𝜑 → ((𝜃𝜓) ↔ (𝜃𝜒)))
 Colors of variables: wff set class Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ↔ wb 104   ∨ wo 697   ⊻ wxo 1353 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 603  ax-in2 604  ax-io 698 This theorem depends on definitions:  df-bi 116  df-xor 1354 This theorem is referenced by:  xorbi12d  1360
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