Theorem example-E5.02a 664

Description: Hypothesis Elimination Example 2.

Assertion

Ref	Expression
example-E5.02a	⊢ (𝑥 = 𝑥₁ → (∀𝑦∃𝑧((𝑦 + 𝑧) = 𝑥 ∨ (𝑥 + 𝑧) = 𝑦) ↔ ∀𝑦∃𝑧((𝑦 + 𝑧) = 𝑥₁ ∨ (𝑥₁ + 𝑧) = 𝑦)))

Distinct variable group:   𝑥,𝑥₁,𝑦,𝑧

Proof of Theorem example-E5.02a

Step	Hyp	Ref	Expression
1		rcp-NDASM1of1 192	. . . . 5 ⊢ (𝑥 = 𝑥₁ → 𝑥 = 𝑥₁)
2	1	subeqr-P5 635	. . . 4 ⊢ (𝑥 = 𝑥₁ → ((𝑦 + 𝑧) = 𝑥 ↔ (𝑦 + 𝑧) = 𝑥₁))
3	1	subaddl-P5 645	. . . . 5 ⊢ (𝑥 = 𝑥₁ → (𝑥 + 𝑧) = (𝑥₁ + 𝑧))
4	3	subeql-P5 632	. . . 4 ⊢ (𝑥 = 𝑥₁ → ((𝑥 + 𝑧) = 𝑦 ↔ (𝑥₁ + 𝑧) = 𝑦))
5	2, 4	subord-P3.43c 350	. . 3 ⊢ (𝑥 = 𝑥₁ → (((𝑦 + 𝑧) = 𝑥 ∨ (𝑥 + 𝑧) = 𝑦) ↔ ((𝑦 + 𝑧) = 𝑥₁ ∨ (𝑥₁ + 𝑧) = 𝑦)))
6	5	subexv-P5 624	. 2 ⊢ (𝑥 = 𝑥₁ → (∃𝑧((𝑦 + 𝑧) = 𝑥 ∨ (𝑥 + 𝑧) = 𝑦) ↔ ∃𝑧((𝑦 + 𝑧) = 𝑥₁ ∨ (𝑥₁ + 𝑧) = 𝑦)))
7	6	suballv-P5 623	1 ⊢ (𝑥 = 𝑥₁ → (∀𝑦∃𝑧((𝑦 + 𝑧) = 𝑥 ∨ (𝑥 + 𝑧) = 𝑦) ↔ ∀𝑦∃𝑧((𝑦 + 𝑧) = 𝑥₁ ∨ (𝑥₁ + 𝑧) = 𝑦)))

Syntax hints:  term-obj 1   + term-add 4   = wff-equals 6  ∀wff-forall 8   → wff-imp 10   ↔ wff-bi 104   ∨ wff-or 144  ∃wff-exists 595

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-addl 23

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)