Theorem example-E5.01a 663

Description: Hypothesis Elimination Example 1.

Assertion

Ref	Expression
example-E5.01a	⊢ (𝑥 = 𝑥₁ → ((𝑡 ⋅ (𝑢 + 𝑥)) = ((𝑡 ⋅ 𝑢) + (𝑡 ⋅ 𝑥)) ↔ (𝑡 ⋅ (𝑢 + 𝑥₁)) = ((𝑡 ⋅ 𝑢) + (𝑡 ⋅ 𝑥₁))))

Distinct variable group:   𝑥,𝑥₁

Proof of Theorem example-E5.01a

Step	Hyp	Ref	Expression
1		rcp-NDASM1of1 192	. . . 4 ⊢ (𝑥 = 𝑥₁ → 𝑥 = 𝑥₁)
2	1	subaddr-P5 646	. . 3 ⊢ (𝑥 = 𝑥₁ → (𝑢 + 𝑥) = (𝑢 + 𝑥₁))
3	2	submultr-P5 650	. 2 ⊢ (𝑥 = 𝑥₁ → (𝑡 ⋅ (𝑢 + 𝑥)) = (𝑡 ⋅ (𝑢 + 𝑥₁)))
4	1	submultr-P5 650	. . 3 ⊢ (𝑥 = 𝑥₁ → (𝑡 ⋅ 𝑥) = (𝑡 ⋅ 𝑥₁))
5	4	subaddr-P5 646	. 2 ⊢ (𝑥 = 𝑥₁ → ((𝑡 ⋅ 𝑢) + (𝑡 ⋅ 𝑥)) = ((𝑡 ⋅ 𝑢) + (𝑡 ⋅ 𝑥₁)))
6	3, 5	subeqd-P5 637	1 ⊢ (𝑥 = 𝑥₁ → ((𝑡 ⋅ (𝑢 + 𝑥)) = ((𝑡 ⋅ 𝑢) + (𝑡 ⋅ 𝑥)) ↔ (𝑡 ⋅ (𝑢 + 𝑥₁)) = ((𝑡 ⋅ 𝑢) + (𝑡 ⋅ 𝑥₁))))

Syntax hints:  term-obj 1   + term-add 4   ⋅ term-mult 5   = wff-equals 6   → wff-imp 10   ↔ wff-bi 104

This theorem was proved from axioms:  ax-L1 11  ax-L2 12  ax-L3 13  ax-MP 14  ax-GEN 15  ax-L4 16  ax-L5 17  ax-L6 18  ax-L7 19  ax-L9-addr 24  ax-L9-multr 26

This theorem depends on definitions:  df-bi-D2.1 107  df-and-D2.2 133  df-or-D2.3 145  df-true-D2.4 155  df-rcp-AND3 161  df-exists-D5.1 596

This theorem is referenced by: (None)