Theorem had1 1402

Description: If the first parameter is true, the half adder is equivalent to the equality of the other two inputs. (Contributed by Mario Carneiro, 4-Sep-2016.)

Assertion

Ref	Expression
had1	⊢ (φ → (hadd(φ, ψ, χ) ↔ (ψ ↔ χ)))

Proof of Theorem had1

Step	Hyp	Ref	Expression
1		hadbi 1387	. . 3 ⊢ (hadd(φ, ψ, χ) ↔ ((φ ↔ ψ) ↔ χ))
2		biass 348	. . 3 ⊢ (((φ ↔ ψ) ↔ χ) ↔ (φ ↔ (ψ ↔ χ)))
3	1, 2	bitri 240	. 2 ⊢ (hadd(φ, ψ, χ) ↔ (φ ↔ (ψ ↔ χ)))
4		id 19	. . . 4 ⊢ (φ → φ)
5		biidd 228	. . . 4 ⊢ (φ → ((ψ ↔ χ) ↔ (ψ ↔ χ)))
6	4, 5	2thd 231	. . 3 ⊢ (φ → (φ ↔ ((ψ ↔ χ) ↔ (ψ ↔ χ))))
7		biass 348	. . 3 ⊢ (((φ ↔ (ψ ↔ χ)) ↔ (ψ ↔ χ)) ↔ (φ ↔ ((ψ ↔ χ) ↔ (ψ ↔ χ))))
8	6, 7	sylibr 203	. 2 ⊢ (φ → ((φ ↔ (ψ ↔ χ)) ↔ (ψ ↔ χ)))
9	3, 8	syl5bb 248	1 ⊢ (φ → (hadd(φ, ψ, χ) ↔ (ψ ↔ χ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176  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:  had0  1403