Theorem xorbi1d 1312

Description: Deduction joining an equivalence and a right operand to form equivalence of exclusive-or. (Contributed by Jim Kingdon, 7-Oct-2018.)

Hypothesis

Ref	Expression
xorbid.1	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
xorbi1d	⊢ (𝜑 → ((𝜓 ⊻ 𝜃) ↔ (𝜒 ⊻ 𝜃)))

Proof of Theorem xorbi1d

Step	Hyp	Ref	Expression
1		xorbid.1	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
2	1	orbi1d 737	. . 3 ⊢ (𝜑 → ((𝜓 ∨ 𝜃) ↔ (𝜒 ∨ 𝜃)))
3	1	anbi1d 452	. . . 4 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) ↔ (𝜒 ∧ 𝜃)))
4	3	notbid 624	. . 3 ⊢ (𝜑 → (¬ (𝜓 ∧ 𝜃) ↔ ¬ (𝜒 ∧ 𝜃)))
5	2, 4	anbi12d 456	. 2 ⊢ (𝜑 → (((𝜓 ∨ 𝜃) ∧ ¬ (𝜓 ∧ 𝜃)) ↔ ((𝜒 ∨ 𝜃) ∧ ¬ (𝜒 ∧ 𝜃))))
6		df-xor 1307	. 2 ⊢ ((𝜓 ⊻ 𝜃) ↔ ((𝜓 ∨ 𝜃) ∧ ¬ (𝜓 ∧ 𝜃)))
7		df-xor 1307	. 2 ⊢ ((𝜒 ⊻ 𝜃) ↔ ((𝜒 ∨ 𝜃) ∧ ¬ (𝜒 ∧ 𝜃)))
8	5, 6, 7	3bitr4g 221	1 ⊢ (𝜑 → ((𝜓 ⊻ 𝜃) ↔ (𝜒 ⊻ 𝜃)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   ↔ wb 103   ∨ wo 661   ⊻ wxo 1306

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in1 576  ax-in2 577  ax-io 662

This theorem depends on definitions:  df-bi 115  df-xor 1307

This theorem is referenced by:  xorbi12d  1313