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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) ↔ (ψχ)))

Proof of Theorem had1
StepHypRef Expression
1 hadbi 1387 . . 3 (hadd(φ, ψ, χ) ↔ ((φψ) ↔ χ))
2 biass 348 . . 3 (((φψ) ↔ χ) ↔ (φ ↔ (ψχ)))
31, 2bitri 240 . 2 (hadd(φ, ψ, χ) ↔ (φ ↔ (ψχ)))
4 id 19 . . . 4 (φφ)
5 biidd 228 . . . 4 (φ → ((ψχ) ↔ (ψχ)))
64, 52thd 231 . . 3 (φ → (φ ↔ ((ψχ) ↔ (ψχ))))
7 biass 348 . . 3 (((φ ↔ (ψχ)) ↔ (ψχ)) ↔ (φ ↔ ((ψχ) ↔ (ψχ))))
86, 7sylibr 203 . 2 (φ → ((φ ↔ (ψχ)) ↔ (ψχ)))
93, 8syl5bb 248 1 (φ → (hadd(φ, ψ, χ) ↔ (ψχ)))
Colors of variables: wff setvar class
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
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