Description: Equality implies bijection. (Contributed by RP, 9-May-2020)
Ref | Expression | ||
---|---|---|---|
Assertion | trcleq12lem | ⊢ ( ( 𝑅 = 𝑆 ∧ 𝐴 = 𝐵 ) → ( ( 𝑅 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ↔ ( 𝑆 ⊆ 𝐵 ∧ ( 𝐵 ∘ 𝐵 ) ⊆ 𝐵 ) ) ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | cleq1lem | ⊢ ( 𝑅 = 𝑆 → ( ( 𝑅 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ↔ ( 𝑆 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ) ) | |
2 | trcleq2lem | ⊢ ( 𝐴 = 𝐵 → ( ( 𝑆 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ↔ ( 𝑆 ⊆ 𝐵 ∧ ( 𝐵 ∘ 𝐵 ) ⊆ 𝐵 ) ) ) | |
3 | 1 2 | sylan9bb | ⊢ ( ( 𝑅 = 𝑆 ∧ 𝐴 = 𝐵 ) → ( ( 𝑅 ⊆ 𝐴 ∧ ( 𝐴 ∘ 𝐴 ) ⊆ 𝐴 ) ↔ ( 𝑆 ⊆ 𝐵 ∧ ( 𝐵 ∘ 𝐵 ) ⊆ 𝐵 ) ) ) |