Theorem xordc1 1442

Description: Exclusive or implies the left proposition is decidable. (Contributed by Jim Kingdon, 12-Mar-2018.)

Assertion

Ref	Expression
xordc1	⊢ ((𝜑 ⊻ 𝜓) → DECID 𝜑)

Proof of Theorem xordc1

Step	Hyp	Ref	Expression
1		andir 831	. . 3 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)) ↔ ((𝜑 ∧ ¬ (𝜑 ∧ 𝜓)) ∨ (𝜓 ∧ ¬ (𝜑 ∧ 𝜓))))
2		simpl 109	. . . 4 ⊢ ((𝜑 ∧ ¬ (𝜑 ∧ 𝜓)) → 𝜑)
3		imnan 701	. . . . . 6 ⊢ ((𝜓 → ¬ 𝜑) ↔ ¬ (𝜓 ∧ 𝜑))
4		ancom 266	. . . . . 6 ⊢ ((𝜑 ∧ 𝜓) ↔ (𝜓 ∧ 𝜑))
5	3, 4	xchbinxr 694	. . . . 5 ⊢ ((𝜓 → ¬ 𝜑) ↔ ¬ (𝜑 ∧ 𝜓))
6		pm3.35 347	. . . . 5 ⊢ ((𝜓 ∧ (𝜓 → ¬ 𝜑)) → ¬ 𝜑)
7	5, 6	sylan2br 288	. . . 4 ⊢ ((𝜓 ∧ ¬ (𝜑 ∧ 𝜓)) → ¬ 𝜑)
8	2, 7	orim12i 771	. . 3 ⊢ (((𝜑 ∧ ¬ (𝜑 ∧ 𝜓)) ∨ (𝜓 ∧ ¬ (𝜑 ∧ 𝜓))) → (𝜑 ∨ ¬ 𝜑))
9	1, 8	sylbi 121	. 2 ⊢ (((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)) → (𝜑 ∨ ¬ 𝜑))
10		df-xor 1425	. 2 ⊢ ((𝜑 ⊻ 𝜓) ↔ ((𝜑 ∨ 𝜓) ∧ ¬ (𝜑 ∧ 𝜓)))
11		df-dc 847	. 2 ⊢ (DECID 𝜑 ↔ (𝜑 ∨ ¬ 𝜑))
12	9, 10, 11	3imtr4i 201	1 ⊢ ((𝜑 ⊻ 𝜓) → DECID 𝜑)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 104   ∨ wo 720  DECID wdc 846   ⊻ wxo 1424

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721

This proof depends on definitions:  df-bi 117  df-dc 847  df-xor 1425

This theorem is used by: (None)