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Mirrors > Home > NFE Home > Th. List > hadbi123i | GIF version |
Description: Equality theorem for half adder. (Contributed by Mario Carneiro, 4-Sep-2016.) |
Ref | Expression |
---|---|
hadbii.1 | ⊢ (φ ↔ ψ) |
hadbii.2 | ⊢ (χ ↔ θ) |
hadbii.3 | ⊢ (τ ↔ η) |
Ref | Expression |
---|---|
hadbi123i | ⊢ (hadd(φ, χ, τ) ↔ hadd(ψ, θ, η)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | hadbii.1 | . . . 4 ⊢ (φ ↔ ψ) | |
2 | 1 | a1i 10 | . . 3 ⊢ ( ⊤ → (φ ↔ ψ)) |
3 | hadbii.2 | . . . 4 ⊢ (χ ↔ θ) | |
4 | 3 | a1i 10 | . . 3 ⊢ ( ⊤ → (χ ↔ θ)) |
5 | hadbii.3 | . . . 4 ⊢ (τ ↔ η) | |
6 | 5 | a1i 10 | . . 3 ⊢ ( ⊤ → (τ ↔ η)) |
7 | 2, 4, 6 | hadbi123d 1382 | . 2 ⊢ ( ⊤ → (hadd(φ, χ, τ) ↔ hadd(ψ, θ, η))) |
8 | 7 | trud 1323 | 1 ⊢ (hadd(φ, χ, τ) ↔ hadd(ψ, θ, η)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 176 ⊤ wtru 1316 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-tru 1319 df-had 1380 |
This theorem is referenced by: (None) |
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