Metamath Proof Explorer


Theorem issubgoilem

Description: Lemma for hhssabloilem . (Contributed by Paul Chapman, 25-Feb-2008) (New usage is discouraged.)

Ref Expression
Hypothesis issubgoilem.1 ⊢ ( ( 𝑥 ∈ 𝑌 ∧ 𝑦 ∈ 𝑌 ) → ( 𝑥 𝐻 𝑦 ) = ( 𝑥 𝐺 𝑦 ) )
Assertion issubgoilem ( ( 𝐴 ∈ 𝑌 ∧ 𝐵 ∈ 𝑌 ) → ( 𝐴 𝐻 𝐵 ) = ( 𝐴 𝐺 𝐵 ) )

Proof

Step Hyp Ref Expression
1 issubgoilem.1 ⊢ ( ( 𝑥 ∈ 𝑌 ∧ 𝑦 ∈ 𝑌 ) → ( 𝑥 𝐻 𝑦 ) = ( 𝑥 𝐺 𝑦 ) )
2 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝐻 𝑦 ) = ( 𝐴 𝐻 𝑦 ) )
3 oveq1 ⊢ ( 𝑥 = 𝐴 → ( 𝑥 𝐺 𝑦 ) = ( 𝐴 𝐺 𝑦 ) )
4 2 3 eqeq12d ⊢ ( 𝑥 = 𝐴 → ( ( 𝑥 𝐻 𝑦 ) = ( 𝑥 𝐺 𝑦 ) ↔ ( 𝐴 𝐻 𝑦 ) = ( 𝐴 𝐺 𝑦 ) ) )
5 oveq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 𝐻 𝑦 ) = ( 𝐴 𝐻 𝐵 ) )
6 oveq2 ⊢ ( 𝑦 = 𝐵 → ( 𝐴 𝐺 𝑦 ) = ( 𝐴 𝐺 𝐵 ) )
7 5 6 eqeq12d ⊢ ( 𝑦 = 𝐵 → ( ( 𝐴 𝐻 𝑦 ) = ( 𝐴 𝐺 𝑦 ) ↔ ( 𝐴 𝐻 𝐵 ) = ( 𝐴 𝐺 𝐵 ) ) )
8 4 7 1 vtocl2ga ⊢ ( ( 𝐴 ∈ 𝑌 ∧ 𝐵 ∈ 𝑌 ) → ( 𝐴 𝐻 𝐵 ) = ( 𝐴 𝐺 𝐵 ) )