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Theorem xoranor 1356
Description: One way of defining exclusive or. Equivalent to df-xor 1355. (Contributed by Jim Kingdon and Mario Carneiro, 1-Mar-2018.)
Assertion
Ref Expression
xoranor ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)))

Proof of Theorem xoranor
StepHypRef Expression
1 df-xor 1355 . . 3 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ ¬ (𝜑𝜓)))
2 ax-ia3 107 . . . . . . 7 (𝜑 → (𝜓 → (𝜑𝜓)))
32con3d 621 . . . . . 6 (𝜑 → (¬ (𝜑𝜓) → ¬ 𝜓))
4 olc 701 . . . . . 6 𝜓 → (¬ 𝜑 ∨ ¬ 𝜓))
53, 4syl6 33 . . . . 5 (𝜑 → (¬ (𝜑𝜓) → (¬ 𝜑 ∨ ¬ 𝜓)))
6 pm3.21 262 . . . . . . 7 (𝜓 → (𝜑 → (𝜑𝜓)))
76con3d 621 . . . . . 6 (𝜓 → (¬ (𝜑𝜓) → ¬ 𝜑))
8 orc 702 . . . . . 6 𝜑 → (¬ 𝜑 ∨ ¬ 𝜓))
97, 8syl6 33 . . . . 5 (𝜓 → (¬ (𝜑𝜓) → (¬ 𝜑 ∨ ¬ 𝜓)))
105, 9jaoi 706 . . . 4 ((𝜑𝜓) → (¬ (𝜑𝜓) → (¬ 𝜑 ∨ ¬ 𝜓)))
1110imdistani 442 . . 3 (((𝜑𝜓) ∧ ¬ (𝜑𝜓)) → ((𝜑𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)))
121, 11sylbi 120 . 2 ((𝜑𝜓) → ((𝜑𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)))
13 pm3.14 743 . . . 4 ((¬ 𝜑 ∨ ¬ 𝜓) → ¬ (𝜑𝜓))
1413anim2i 340 . . 3 (((𝜑𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)) → ((𝜑𝜓) ∧ ¬ (𝜑𝜓)))
1514, 1sylibr 133 . 2 (((𝜑𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)) → (𝜑𝜓))
1612, 15impbii 125 1 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (¬ 𝜑 ∨ ¬ 𝜓)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 103  wb 104  wo 698  wxo 1354
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 604  ax-in2 605  ax-io 699
This theorem depends on definitions:  df-bi 116  df-xor 1355
This theorem is referenced by:  excxor  1357  xoror  1358
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