Theorem dcan 929

Description: A conjunction of two decidable propositions is decidable. (Contributed by Jim Kingdon, 12-Apr-2018.)

Assertion

Ref	Expression
dcan	⊢ ((DECID 𝜑 ∧ DECID 𝜓) → DECID (𝜑 ∧ 𝜓))

Proof of Theorem dcan

Step	Hyp	Ref	Expression
1		simpl 108	. . . . 5 ⊢ ((¬ 𝜑 ∧ 𝜓) → ¬ 𝜑)
2	1	intnanrd 928	. . . 4 ⊢ ((¬ 𝜑 ∧ 𝜓) → ¬ (𝜑 ∧ 𝜓))
3	2	orim2i 757	. . 3 ⊢ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)) → ((𝜑 ∧ 𝜓) ∨ ¬ (𝜑 ∧ 𝜓)))
4		simpr 109	. . . . 5 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓) → ¬ 𝜓)
5	4	intnand 927	. . . 4 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓) → ¬ (𝜑 ∧ 𝜓))
6	5	olcd 730	. . 3 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓) → ((𝜑 ∧ 𝜓) ∨ ¬ (𝜑 ∧ 𝜓)))
7	3, 6	jaoi 712	. 2 ⊢ ((((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)) → ((𝜑 ∧ 𝜓) ∨ ¬ (𝜑 ∧ 𝜓)))
8		df-dc 831	. . . 4 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑))
9		df-dc 831	. . . 4 ⊢ (DECID 𝜓 ↔ (𝜓 ∨ ¬ 𝜓))
10	8, 9	anbi12i 458	. . 3 ⊢ ((DECID 𝜑 ∧ DECID 𝜓) ↔ ((𝜑 ∨ ¬ 𝜑) ∧ (𝜓 ∨ ¬ 𝜓)))
11		andi 814	. . 3 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ (𝜓 ∨ ¬ 𝜓)) ↔ (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)))
12		andir 815	. . . 4 ⊢ (((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ↔ ((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)))
13	12	orbi1i 759	. . 3 ⊢ ((((𝜑 ∨ ¬ 𝜑) ∧ 𝜓) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)) ↔ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)))
14	10, 11, 13	3bitri 205	. 2 ⊢ ((DECID 𝜑 ∧ DECID 𝜓) ↔ (((𝜑 ∧ 𝜓) ∨ (¬ 𝜑 ∧ 𝜓)) ∨ ((𝜑 ∨ ¬ 𝜑) ∧ ¬ 𝜓)))
15		df-dc 831	. 2 ⊢ (DECID (𝜑 ∧ 𝜓) ↔ ((𝜑 ∧ 𝜓) ∨ ¬ (𝜑 ∧ 𝜓)))
16	7, 14, 15	3imtr4i 200	1 ⊢ ((DECID 𝜑 ∧ DECID 𝜓) → DECID (𝜑 ∧ 𝜓))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 103   ∨ wo 704  DECID wdc 830

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 610  ax-in2 611  ax-io 705

This theorem depends on definitions:  df-bi 116  df-dc 831

This theorem is referenced by:  dcan2  930