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Metamath Proof Explorer

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 ) )