Theorem nnacan

Description: Cancellation law for addition of natural numbers. (Contributed by NM, 27-Oct-1995) (Revised by Mario Carneiro, 15-Nov-2014)

	Ref	Expression
Assertion	nnacan	⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) → ( ( 𝐴 +o 𝐵 ) = ( 𝐴 +o 𝐶 ) ↔ 𝐵 = 𝐶 ) )

Proof

Step	Hyp	Ref	Expression
1		nnaword	⊢ ( ( 𝐵 ∈ ω ∧ 𝐶 ∈ ω ∧ 𝐴 ∈ ω ) → ( 𝐵 ⊆ 𝐶 ↔ ( 𝐴 +o 𝐵 ) ⊆ ( 𝐴 +o 𝐶 ) ) )
2	1	3comr	⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) → ( 𝐵 ⊆ 𝐶 ↔ ( 𝐴 +o 𝐵 ) ⊆ ( 𝐴 +o 𝐶 ) ) )
3		nnaword	⊢ ( ( 𝐶 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐴 ∈ ω ) → ( 𝐶 ⊆ 𝐵 ↔ ( 𝐴 +o 𝐶 ) ⊆ ( 𝐴 +o 𝐵 ) ) )
4	3	3com13	⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) → ( 𝐶 ⊆ 𝐵 ↔ ( 𝐴 +o 𝐶 ) ⊆ ( 𝐴 +o 𝐵 ) ) )
5	2 4	anbi12d	⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) → ( ( 𝐵 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵 ) ↔ ( ( 𝐴 +o 𝐵 ) ⊆ ( 𝐴 +o 𝐶 ) ∧ ( 𝐴 +o 𝐶 ) ⊆ ( 𝐴 +o 𝐵 ) ) ) )
6	5	bicomd	⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) → ( ( ( 𝐴 +o 𝐵 ) ⊆ ( 𝐴 +o 𝐶 ) ∧ ( 𝐴 +o 𝐶 ) ⊆ ( 𝐴 +o 𝐵 ) ) ↔ ( 𝐵 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵 ) ) )
7		eqss	⊢ ( ( 𝐴 +o 𝐵 ) = ( 𝐴 +o 𝐶 ) ↔ ( ( 𝐴 +o 𝐵 ) ⊆ ( 𝐴 +o 𝐶 ) ∧ ( 𝐴 +o 𝐶 ) ⊆ ( 𝐴 +o 𝐵 ) ) )
8		eqss	⊢ ( 𝐵 = 𝐶 ↔ ( 𝐵 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵 ) )
9	6 7 8	3bitr4g	⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) → ( ( 𝐴 +o 𝐵 ) = ( 𝐴 +o 𝐶 ) ↔ 𝐵 = 𝐶 ) )