Theorem xorbi1d 1270

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 704	. . 3 ⊢ (φ → ((ψ ∨ θ) ↔ (χ ∨ θ)))
3	1	anbi1d 438	. . . 4 ⊢ (φ → ((ψ ∧ θ) ↔ (χ ∧ θ)))
4	3	notbid 591	. . 3 ⊢ (φ → (¬ (ψ ∧ θ) ↔ ¬ (χ ∧ θ)))
5	2, 4	anbi12d 442	. 2 ⊢ (φ → (((ψ ∨ θ) ∧ ¬ (ψ ∧ θ)) ↔ ((χ ∨ θ) ∧ ¬ (χ ∧ θ))))
6		df-xor 1266	. 2 ⊢ ((ψ ⊻ θ) ↔ ((ψ ∨ θ) ∧ ¬ (ψ ∧ θ)))
7		df-xor 1266	. 2 ⊢ ((χ ⊻ θ) ↔ ((χ ∨ θ) ∧ ¬ (χ ∧ θ)))
8	5, 6, 7	3bitr4g 212	1 ⊢ (φ → ((ψ ⊻ θ) ↔ (χ ⊻ θ)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 97   ↔ wb 98   ∨ wo 628   ⊻ wxo 1265

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-in1 544  ax-in2 545  ax-io 629

This theorem depends on definitions:  df-bi 110  df-xor 1266

This theorem is referenced by:  xorbi12d  1271