Theorem cad0 1400

Description: If one parameter is false, the adder carry is true exactly when both of the other two parameters are true. (Contributed by Mario Carneiro, 8-Sep-2016.)

Assertion

Ref	Expression
cad0	⊢ (¬ χ → (cadd(φ, ψ, χ) ↔ (φ ∧ ψ)))

Proof of Theorem cad0

Step	Hyp	Ref	Expression
1		df-cad 1381	. 2 ⊢ (cadd(φ, ψ, χ) ↔ ((φ ∧ ψ) ∨ (χ ∧ (φ ⊻ ψ))))
2		idd 21	. . . 4 ⊢ (¬ χ → ((φ ∧ ψ) → (φ ∧ ψ)))
3		pm2.21 100	. . . . 5 ⊢ (¬ χ → (χ → (φ ∧ ψ)))
4	3	adantrd 454	. . . 4 ⊢ (¬ χ → ((χ ∧ (φ ⊻ ψ)) → (φ ∧ ψ)))
5	2, 4	jaod 369	. . 3 ⊢ (¬ χ → (((φ ∧ ψ) ∨ (χ ∧ (φ ⊻ ψ))) → (φ ∧ ψ)))
6		orc 374	. . 3 ⊢ ((φ ∧ ψ) → ((φ ∧ ψ) ∨ (χ ∧ (φ ⊻ ψ))))
7	5, 6	impbid1 194	. 2 ⊢ (¬ χ → (((φ ∧ ψ) ∨ (χ ∧ (φ ⊻ ψ))) ↔ (φ ∧ ψ)))
8	1, 7	syl5bb 248	1 ⊢ (¬ χ → (cadd(φ, ψ, χ) ↔ (φ ∧ ψ)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 176   ∨ wo 357   ∧ wa 358   ⊻ wxo 1304  caddwcad 1379

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-or 359  df-an 360  df-cad 1381

This theorem is used by: (None)