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Metamath Proof Explorer

Theorem nnmcan

Description: Cancellation law for multiplication of natural numbers. (Contributed by NM, 26-Oct-1995) (Revised by Mario Carneiro, 15-Nov-2014)

Ref Expression
Assertion nnmcan ( ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( ( 𝐴 ·o 𝐵 ) = ( 𝐴 ·o 𝐶 ) ↔ 𝐵 = 𝐶 ) )

Proof

Step Hyp Ref Expression
1 3anrot ⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ↔ ( 𝐵 ∈ ω ∧ 𝐶 ∈ ω ∧ 𝐴 ∈ ω ) )
2 nnmword ⊢ ( ( ( 𝐵 ∈ ω ∧ 𝐶 ∈ ω ∧ 𝐴 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( 𝐵 ⊆ 𝐶 ↔ ( 𝐴 ·o 𝐵 ) ⊆ ( 𝐴 ·o 𝐶 ) ) )
3 1 2 sylanb ⊢ ( ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( 𝐵 ⊆ 𝐶 ↔ ( 𝐴 ·o 𝐵 ) ⊆ ( 𝐴 ·o 𝐶 ) ) )
4 3anrev ⊢ ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ↔ ( 𝐶 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐴 ∈ ω ) )
5 nnmword ⊢ ( ( ( 𝐶 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐴 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( 𝐶 ⊆ 𝐵 ↔ ( 𝐴 ·o 𝐶 ) ⊆ ( 𝐴 ·o 𝐵 ) ) )
6 4 5 sylanb ⊢ ( ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( 𝐶 ⊆ 𝐵 ↔ ( 𝐴 ·o 𝐶 ) ⊆ ( 𝐴 ·o 𝐵 ) ) )
7 3 6 anbi12d ⊢ ( ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( ( 𝐵 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵 ) ↔ ( ( 𝐴 ·o 𝐵 ) ⊆ ( 𝐴 ·o 𝐶 ) ∧ ( 𝐴 ·o 𝐶 ) ⊆ ( 𝐴 ·o 𝐵 ) ) ) )
8 7 bicomd ⊢ ( ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( ( ( 𝐴 ·o 𝐵 ) ⊆ ( 𝐴 ·o 𝐶 ) ∧ ( 𝐴 ·o 𝐶 ) ⊆ ( 𝐴 ·o 𝐵 ) ) ↔ ( 𝐵 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵 ) ) )
9 eqss ⊢ ( ( 𝐴 ·o 𝐵 ) = ( 𝐴 ·o 𝐶 ) ↔ ( ( 𝐴 ·o 𝐵 ) ⊆ ( 𝐴 ·o 𝐶 ) ∧ ( 𝐴 ·o 𝐶 ) ⊆ ( 𝐴 ·o 𝐵 ) ) )
10 eqss ⊢ ( 𝐵 = 𝐶 ↔ ( 𝐵 ⊆ 𝐶 ∧ 𝐶 ⊆ 𝐵 ) )
11 8 9 10 3bitr4g ⊢ ( ( ( 𝐴 ∈ ω ∧ 𝐵 ∈ ω ∧ 𝐶 ∈ ω ) ∧ ∅ ∈ 𝐴 ) → ( ( 𝐴 ·o 𝐵 ) = ( 𝐴 ·o 𝐶 ) ↔ 𝐵 = 𝐶 ) )