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Theorem dcan 851
Description: A conjunction of two decidable propositions is decidable. (Contributed by Jim Kingdon, 12-Apr-2018.)
Assertion
Ref Expression
dcan (DECID 𝜑 → (DECID 𝜓DECID (𝜑𝜓)))

Proof of Theorem dcan
StepHypRef Expression
1 simpl 106 . . . . . 6 ((¬ 𝜑𝜓) → ¬ 𝜑)
21intnanrd 850 . . . . 5 ((¬ 𝜑𝜓) → ¬ (𝜑𝜓))
32orim2i 686 . . . 4 (((𝜑𝜓) ∨ (¬ 𝜑𝜓)) → ((𝜑𝜓) ∨ ¬ (𝜑𝜓)))
4 simpr 107 . . . . . 6 (((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓) → ¬ 𝜓)
54intnand 849 . . . . 5 (((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓) → ¬ (𝜑𝜓))
65olcd 661 . . . 4 (((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓) → ((𝜑𝜓) ∨ ¬ (𝜑𝜓)))
73, 6jaoi 644 . . 3 ((((𝜑𝜓) ∨ (¬ 𝜑𝜓)) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)) → ((𝜑𝜓) ∨ ¬ (𝜑𝜓)))
8 df-dc 752 . . . . 5 (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑))
9 df-dc 752 . . . . 5 (DECID 𝜓 ↔ (𝜓 ∨ ¬ 𝜓))
108, 9anbi12i 441 . . . 4 ((DECID 𝜑DECID 𝜓) ↔ ((𝜑 ∨ ¬ 𝜑) ∧ (𝜓 ∨ ¬ 𝜓)))
11 andi 740 . . . 4 (((𝜑 ∨ ¬ 𝜑) ∧ (𝜓 ∨ ¬ 𝜓)) ↔ (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)))
12 andir 741 . . . . 5 (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ↔ ((𝜑𝜓) ∨ (¬ 𝜑𝜓)))
1312orbi1i 688 . . . 4 ((((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)) ↔ (((𝜑𝜓) ∨ (¬ 𝜑𝜓)) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)))
1410, 11, 133bitri 199 . . 3 ((DECID 𝜑DECID 𝜓) ↔ (((𝜑𝜓) ∨ (¬ 𝜑𝜓)) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)))
15 df-dc 752 . . 3 (DECID (𝜑𝜓) ↔ ((𝜑𝜓) ∨ ¬ (𝜑𝜓)))
167, 14, 153imtr4i 194 . 2 ((DECID 𝜑DECID 𝜓) → DECID (𝜑𝜓))
1716ex 112 1 (DECID 𝜑 → (DECID 𝜓DECID (𝜑𝜓)))
Colors of variables: wff set class
Syntax hints:  ¬ wn 3  wi 4  wa 101  wo 637  DECID wdc 751
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 552  ax-in2 553  ax-io 638
This theorem depends on definitions:  df-bi 114  df-dc 752
This theorem is referenced by:  dcbi  853  annimdc  854  pm4.55dc  855  anordc  872  xordidc  1304  nn0n0n1ge2b  8348
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