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	|- (ph -> (hadd(ph, ps, ch) <-> (ps <-> ch)))

Proof of Theorem had1

Step	Hyp	Ref	Expression
1		hadbi 1387	. . 3 |- (hadd(ph, ps, ch) <-> ((ph <-> ps) <-> ch))
2		biass 348	. . 3 |- (((ph <-> ps) <-> ch) <-> (ph <-> (ps <-> ch)))
3	1, 2	bitri 240	. 2 |- (hadd(ph, ps, ch) <-> (ph <-> (ps <-> ch)))
4		id 19	. . . 4 |- (ph -> ph)
5		biidd 228	. . . 4 |- (ph -> ((ps <-> ch) <-> (ps <-> ch)))
6	4, 5	2thd 231	. . 3 |- (ph -> (ph <-> ((ps <-> ch) <-> (ps <-> ch))))
7		biass 348	. . 3 |- (((ph <-> (ps <-> ch)) <-> (ps <-> ch)) <-> (ph <-> ((ps <-> ch) <-> (ps <-> ch))))
8	6, 7	sylibr 203	. 2 |- (ph -> ((ph <-> (ps <-> ch)) <-> (ps <-> ch)))
9	3, 8	syl5bb 248	1 |- (ph -> (hadd(ph, ps, ch) <-> (ps <-> ch)))

Syntax hints:   -> wi 4   <-> wb 176  haddwhad 1378

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

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

This theorem is referenced by:  had0  1403