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 ) ) ) )