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Metamath Proof Explorer

Theorem swopolem

Description: Perform the substitutions into the strict weak ordering law. (Contributed by Mario Carneiro, 31-Dec-2014)

Ref Expression
Hypothesis swopolem.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) → ( 𝑥 𝑅 𝑦 → ( 𝑥 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) )
Assertion swopolem ( ( 𝜑 ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ∧ 𝑍 ∈ 𝐴 ) ) → ( 𝑋 𝑅 𝑌 → ( 𝑋 𝑅 𝑍 ∨ 𝑍 𝑅 𝑌 ) ) )

Proof

Step Hyp Ref Expression
1 swopolem.1 ⊢ ( ( 𝜑 ∧ ( 𝑥 ∈ 𝐴 ∧ 𝑦 ∈ 𝐴 ∧ 𝑧 ∈ 𝐴 ) ) → ( 𝑥 𝑅 𝑦 → ( 𝑥 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) )
2 1 ralrimivvva ⊢ ( 𝜑 → ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( 𝑥 𝑅 𝑦 → ( 𝑥 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) )
3 breq1 ⊢ ( 𝑥 = 𝑋 → ( 𝑥 𝑅 𝑦 ↔ 𝑋 𝑅 𝑦 ) )
4 breq1 ⊢ ( 𝑥 = 𝑋 → ( 𝑥 𝑅 𝑧 ↔ 𝑋 𝑅 𝑧 ) )
5 4 orbi1d ⊢ ( 𝑥 = 𝑋 → ( ( 𝑥 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ↔ ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) )
6 3 5 imbi12d ⊢ ( 𝑥 = 𝑋 → ( ( 𝑥 𝑅 𝑦 → ( 𝑥 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) ↔ ( 𝑋 𝑅 𝑦 → ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) ) )
7 breq2 ⊢ ( 𝑦 = 𝑌 → ( 𝑋 𝑅 𝑦 ↔ 𝑋 𝑅 𝑌 ) )
8 breq2 ⊢ ( 𝑦 = 𝑌 → ( 𝑧 𝑅 𝑦 ↔ 𝑧 𝑅 𝑌 ) )
9 8 orbi2d ⊢ ( 𝑦 = 𝑌 → ( ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ↔ ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑌 ) ) )
10 7 9 imbi12d ⊢ ( 𝑦 = 𝑌 → ( ( 𝑋 𝑅 𝑦 → ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) ↔ ( 𝑋 𝑅 𝑌 → ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑌 ) ) ) )
11 breq2 ⊢ ( 𝑧 = 𝑍 → ( 𝑋 𝑅 𝑧 ↔ 𝑋 𝑅 𝑍 ) )
12 breq1 ⊢ ( 𝑧 = 𝑍 → ( 𝑧 𝑅 𝑌 ↔ 𝑍 𝑅 𝑌 ) )
13 11 12 orbi12d ⊢ ( 𝑧 = 𝑍 → ( ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑌 ) ↔ ( 𝑋 𝑅 𝑍 ∨ 𝑍 𝑅 𝑌 ) ) )
14 13 imbi2d ⊢ ( 𝑧 = 𝑍 → ( ( 𝑋 𝑅 𝑌 → ( 𝑋 𝑅 𝑧 ∨ 𝑧 𝑅 𝑌 ) ) ↔ ( 𝑋 𝑅 𝑌 → ( 𝑋 𝑅 𝑍 ∨ 𝑍 𝑅 𝑌 ) ) ) )
15 6 10 14 rspc3v ⊢ ( ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ∧ 𝑍 ∈ 𝐴 ) → ( ∀ 𝑥 ∈ 𝐴 ∀ 𝑦 ∈ 𝐴 ∀ 𝑧 ∈ 𝐴 ( 𝑥 𝑅 𝑦 → ( 𝑥 𝑅 𝑧 ∨ 𝑧 𝑅 𝑦 ) ) → ( 𝑋 𝑅 𝑌 → ( 𝑋 𝑅 𝑍 ∨ 𝑍 𝑅 𝑌 ) ) ) )
16 2 15 mpan9 ⊢ ( ( 𝜑 ∧ ( 𝑋 ∈ 𝐴 ∧ 𝑌 ∈ 𝐴 ∧ 𝑍 ∈ 𝐴 ) ) → ( 𝑋 𝑅 𝑌 → ( 𝑋 𝑅 𝑍 ∨ 𝑍 𝑅 𝑌 ) ) )