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Metamath Proof Explorer

Theorem psr0

Description: The zero element of the ring of power series. (Contributed by Mario Carneiro, 29-Dec-2014)

Ref Expression
Hypotheses psrgrp.s ⊢ 𝑆 = ( 𝐼 mPwSer 𝑅 )
psrgrp.i ⊢ ( 𝜑 → 𝐼 ∈ 𝑉 )
psrgrp.r ⊢ ( 𝜑 → 𝑅 ∈ Grp )
psr0.d ⊢ 𝐷 = { 𝑓 ∈ ( ℕ0 ↑m 𝐼 ) ∣ ( ◡ 𝑓 “ ℕ ) ∈ Fin }
psr0.o ⊢ 𝑂 = ( 0g ‘ 𝑅 )
psr0.z ⊢ 0 = ( 0g ‘ 𝑆 )
Assertion psr0 ( 𝜑 → 0 = ( 𝐷 × { 𝑂 } ) )

Proof

Step Hyp Ref Expression
1 psrgrp.s ⊢ 𝑆 = ( 𝐼 mPwSer 𝑅 )
2 psrgrp.i ⊢ ( 𝜑 → 𝐼 ∈ 𝑉 )
3 psrgrp.r ⊢ ( 𝜑 → 𝑅 ∈ Grp )
4 psr0.d ⊢ 𝐷 = { 𝑓 ∈ ( ℕ0 ↑m 𝐼 ) ∣ ( ◡ 𝑓 “ ℕ ) ∈ Fin }
5 psr0.o ⊢ 𝑂 = ( 0g ‘ 𝑅 )
6 psr0.z ⊢ 0 = ( 0g ‘ 𝑆 )
7 eqid ⊢ ( Base ‘ 𝑆 ) = ( Base ‘ 𝑆 )
8 eqid ⊢ ( +g ‘ 𝑆 ) = ( +g ‘ 𝑆 )
9 1 2 3 4 5 7 psr0cl ⊢ ( 𝜑 → ( 𝐷 × { 𝑂 } ) ∈ ( Base ‘ 𝑆 ) )
10 1 2 3 4 5 7 8 9 psr0lid ⊢ ( 𝜑 → ( ( 𝐷 × { 𝑂 } ) ( +g ‘ 𝑆 ) ( 𝐷 × { 𝑂 } ) ) = ( 𝐷 × { 𝑂 } ) )
11 1 2 3 psrgrp ⊢ ( 𝜑 → 𝑆 ∈ Grp )
12 7 8 6 grpid ⊢ ( ( 𝑆 ∈ Grp ∧ ( 𝐷 × { 𝑂 } ) ∈ ( Base ‘ 𝑆 ) ) → ( ( ( 𝐷 × { 𝑂 } ) ( +g ‘ 𝑆 ) ( 𝐷 × { 𝑂 } ) ) = ( 𝐷 × { 𝑂 } ) ↔ 0 = ( 𝐷 × { 𝑂 } ) ) )
13 11 9 12 syl2anc ⊢ ( 𝜑 → ( ( ( 𝐷 × { 𝑂 } ) ( +g ‘ 𝑆 ) ( 𝐷 × { 𝑂 } ) ) = ( 𝐷 × { 𝑂 } ) ↔ 0 = ( 𝐷 × { 𝑂 } ) ) )
14 10 13 mpbid ⊢ ( 𝜑 → 0 = ( 𝐷 × { 𝑂 } ) )