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Metamath Proof Explorer

Theorem submaval0

Description: Second substitution for a submatrix. (Contributed by AV, 28-Dec-2018)

Ref Expression
Hypotheses submafval.a ⊢ 𝐴 = ( 𝑁 Mat 𝑅 )
submafval.q ⊢ 𝑄 = ( 𝑁 subMat 𝑅 )
submafval.b ⊢ 𝐵 = ( Base ‘ 𝐴 )
Assertion submaval0 ( 𝑀 ∈ 𝐵 → ( 𝑄 ‘ 𝑀 ) = ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) )

Proof

Step Hyp Ref Expression
1 submafval.a ⊢ 𝐴 = ( 𝑁 Mat 𝑅 )
2 submafval.q ⊢ 𝑄 = ( 𝑁 subMat 𝑅 )
3 submafval.b ⊢ 𝐵 = ( Base ‘ 𝐴 )
4 1 3 matrcl ⊢ ( 𝑀 ∈ 𝐵 → ( 𝑁 ∈ Fin ∧ 𝑅 ∈ V ) )
5 4 simpld ⊢ ( 𝑀 ∈ 𝐵 → 𝑁 ∈ Fin )
6 mpoexga ⊢ ( ( 𝑁 ∈ Fin ∧ 𝑁 ∈ Fin ) → ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) ∈ V )
7 5 5 6 syl2anc ⊢ ( 𝑀 ∈ 𝐵 → ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) ∈ V )
8 oveq ⊢ ( 𝑚 = 𝑀 → ( 𝑖 𝑚 𝑗 ) = ( 𝑖 𝑀 𝑗 ) )
9 8 mpoeq3dv ⊢ ( 𝑚 = 𝑀 → ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑚 𝑗 ) ) = ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) )
10 9 mpoeq3dv ⊢ ( 𝑚 = 𝑀 → ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑚 𝑗 ) ) ) = ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) )
11 1 2 3 submafval ⊢ 𝑄 = ( 𝑚 ∈ 𝐵 ↦ ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑚 𝑗 ) ) ) )
12 10 11 fvmptg ⊢ ( ( 𝑀 ∈ 𝐵 ∧ ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) ∈ V ) → ( 𝑄 ‘ 𝑀 ) = ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) )
13 7 12 mpdan ⊢ ( 𝑀 ∈ 𝐵 → ( 𝑄 ‘ 𝑀 ) = ( 𝑘 ∈ 𝑁 , 𝑙 ∈ 𝑁 ↦ ( 𝑖 ∈ ( 𝑁 ∖ { 𝑘 } ) , 𝑗 ∈ ( 𝑁 ∖ { 𝑙 } ) ↦ ( 𝑖 𝑀 𝑗 ) ) ) )