Metamath Proof Explorer


Theorem psseq2

Description: Equality theorem for proper subclass. (Contributed by NM, 7-Feb-1996)

Ref Expression
Assertion psseq2 ( 𝐴 = 𝐵 → ( 𝐶 ⊊ 𝐴 ↔ 𝐶 ⊊ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 sseq2 ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ⊆ 𝐴 ↔ 𝐶 ⊆ 𝐵 ) )
2 neeq2 ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ≠ 𝐴 ↔ 𝐶 ≠ 𝐵 ) )
3 1 2 anbi12d ⊢ ( 𝐴 = 𝐵 → ( ( 𝐶 ⊆ 𝐴 ∧ 𝐶 ≠ 𝐴 ) ↔ ( 𝐶 ⊆ 𝐵 ∧ 𝐶 ≠ 𝐵 ) ) )
4 df-pss ⊢ ( 𝐶 ⊊ 𝐴 ↔ ( 𝐶 ⊆ 𝐴 ∧ 𝐶 ≠ 𝐴 ) )
5 df-pss ⊢ ( 𝐶 ⊊ 𝐵 ↔ ( 𝐶 ⊆ 𝐵 ∧ 𝐶 ≠ 𝐵 ) )
6 3 4 5 3bitr4g ⊢ ( 𝐴 = 𝐵 → ( 𝐶 ⊊ 𝐴 ↔ 𝐶 ⊊ 𝐵 ) )