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Metamath Proof Explorer

Theorem decadd

Description: Add two numerals M and N (no carry). (Contributed by Mario Carneiro, 18-Feb-2014) (Revised by AV, 6-Sep-2021)

Ref Expression
Hypotheses decma.a ⊢ 𝐴 ∈ ℕ0
decma.b ⊢ 𝐵 ∈ ℕ0
decma.c ⊢ 𝐶 ∈ ℕ0
decma.d ⊢ 𝐷 ∈ ℕ0
decma.m ⊢ 𝑀 = 𝐴 𝐵
decma.n ⊢ 𝑁 = 𝐶 𝐷
decadd.e ⊢ ( 𝐴 + 𝐶 ) = 𝐸
decadd.f ⊢ ( 𝐵 + 𝐷 ) = 𝐹
Assertion decadd ( 𝑀 + 𝑁 ) = 𝐸 𝐹

Proof

Step Hyp Ref Expression
1 decma.a ⊢ 𝐴 ∈ ℕ0
2 decma.b ⊢ 𝐵 ∈ ℕ0
3 decma.c ⊢ 𝐶 ∈ ℕ0
4 decma.d ⊢ 𝐷 ∈ ℕ0
5 decma.m ⊢ 𝑀 = 𝐴 𝐵
6 decma.n ⊢ 𝑁 = 𝐶 𝐷
7 decadd.e ⊢ ( 𝐴 + 𝐶 ) = 𝐸
8 decadd.f ⊢ ( 𝐵 + 𝐷 ) = 𝐹
9 10nn0 ⊢ 1 0 ∈ ℕ0
10 dfdec10 ⊢ 𝐴 𝐵 = ( ( 1 0 · 𝐴 ) + 𝐵 )
11 5 10 eqtri ⊢ 𝑀 = ( ( 1 0 · 𝐴 ) + 𝐵 )
12 dfdec10 ⊢ 𝐶 𝐷 = ( ( 1 0 · 𝐶 ) + 𝐷 )
13 6 12 eqtri ⊢ 𝑁 = ( ( 1 0 · 𝐶 ) + 𝐷 )
14 9 1 2 3 4 11 13 7 8 numadd ⊢ ( 𝑀 + 𝑁 ) = ( ( 1 0 · 𝐸 ) + 𝐹 )
15 dfdec10 ⊢ 𝐸 𝐹 = ( ( 1 0 · 𝐸 ) + 𝐹 )
16 14 15 eqtr4i ⊢ ( 𝑀 + 𝑁 ) = 𝐸 𝐹