Theorem caselem 561

Description: Lemma for combining cases.

Assertion

Ref	Expression
caselem	⊢ (((φ ∨ ψ) ∧ (χ ∨ θ)) ↔ (((φ ∧ χ) ∨ (ψ ∧ χ)) ∨ ((φ ∧ θ) ∨ (ψ ∧ θ))))

Proof of Theorem caselem

Step	Hyp	Ref	Expression
1		andi 456	. 2 ⊢ (((φ ∨ ψ) ∧ (χ ∨ θ)) ↔ (((φ ∨ ψ) ∧ χ) ∨ ((φ ∨ ψ) ∧ θ)))
2		andir 457	. . 3 ⊢ (((φ ∨ ψ) ∧ χ) ↔ ((φ ∧ χ) ∨ (ψ ∧ χ)))
3		andir 457	. . 3 ⊢ (((φ ∨ ψ) ∧ θ) ↔ ((φ ∧ θ) ∨ (ψ ∧ θ)))
4	2, 3	orbi12i 216	. 2 ⊢ ((((φ ∨ ψ) ∧ χ) ∨ ((φ ∨ ψ) ∧ θ)) ↔ (((φ ∧ χ) ∨ (ψ ∧ χ)) ∨ ((φ ∧ θ) ∨ (ψ ∧ θ))))
5	1, 4	bitr 151	1 ⊢ (((φ ∨ ψ) ∧ (χ ∨ θ)) ↔ (((φ ∧ χ) ∨ (ψ ∧ χ)) ∨ ((φ ∧ θ) ∨ (ψ ∧ θ))))

Syntax hints:   ↔ wb 127   ∨ wo 195   ∧ wa 196

This theorem is referenced by:  ccase 562  ccased 563

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

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