Theorem hadbi123d 1382

Description: Equality theorem for half adder. (Contributed by Mario Carneiro, 4-Sep-2016.)

Hypotheses

Ref	Expression
hadbid.1	⊢ (φ → (ψ ↔ χ))
hadbid.2	⊢ (φ → (θ ↔ τ))
hadbid.3	⊢ (φ → (η ↔ ζ))

Assertion

Ref	Expression
hadbi123d	⊢ (φ → (hadd(ψ, θ, η) ↔ hadd(χ, τ, ζ)))

Proof of Theorem hadbi123d

Step	Hyp	Ref	Expression
1		hadbid.1	. . . 4 ⊢ (φ → (ψ ↔ χ))
2		hadbid.2	. . . 4 ⊢ (φ → (θ ↔ τ))
3	1, 2	xorbi12d 1315	. . 3 ⊢ (φ → ((ψ ⊻ θ) ↔ (χ ⊻ τ)))
4		hadbid.3	. . 3 ⊢ (φ → (η ↔ ζ))
5	3, 4	xorbi12d 1315	. 2 ⊢ (φ → (((ψ ⊻ θ) ⊻ η) ↔ ((χ ⊻ τ) ⊻ ζ)))
6		df-had 1380	. 2 ⊢ (hadd(ψ, θ, η) ↔ ((ψ ⊻ θ) ⊻ η))
7		df-had 1380	. 2 ⊢ (hadd(χ, τ, ζ) ↔ ((χ ⊻ τ) ⊻ ζ))
8	5, 6, 7	3bitr4g 279	1 ⊢ (φ → (hadd(ψ, θ, η) ↔ hadd(χ, τ, ζ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ⊻ wxo 1304  haddwhad 1378

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 177  df-xor 1305  df-had 1380

This theorem is used by:  hadbi123i  1384