Theorem xorbi1d 1430

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 803	. . 3 ⊢ (𝜑 → ((𝜓 ∨ 𝜃) ↔ (𝜒 ∨ 𝜃)))
3	1	anbi1d 469	. . . 4 ⊢ (𝜑 → ((𝜓 ∧ 𝜃) ↔ (𝜒 ∧ 𝜃)))
4	3	notbid 677	. . 3 ⊢ (𝜑 → (¬ (𝜓 ∧ 𝜃) ↔ ¬ (𝜒 ∧ 𝜃)))
5	2, 4	anbi12d 477	. 2 ⊢ (𝜑 → (((𝜓 ∨ 𝜃) ∧ ¬ (𝜓 ∧ 𝜃)) ↔ ((𝜒 ∨ 𝜃) ∧ ¬ (𝜒 ∧ 𝜃))))
6		df-xor 1425	. 2 ⊢ ((𝜓 ⊻ 𝜃) ↔ ((𝜓 ∨ 𝜃) ∧ ¬ (𝜓 ∧ 𝜃)))
7		df-xor 1425	. 2 ⊢ ((𝜒 ⊻ 𝜃) ↔ ((𝜒 ∨ 𝜃) ∧ ¬ (𝜒 ∧ 𝜃)))
8	5, 6, 7	3bitr4g 223	1 ⊢ (𝜑 → ((𝜓 ⊻ 𝜃) ↔ (𝜒 ⊻ 𝜃)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   ∨ wo 720   ⊻ 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-xor 1425

This theorem is used by:  xorbi12d  1431