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Metamath Proof Explorer

Theorem ichbidv

Description: Formula building rule for interchangeability (deduction). (Contributed by SN, 4-May-2024)

Ref Expression
Hypothesis ichbidv.1 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
Assertion ichbidv ( 𝜑 → ( [ 𝑥 ⇄ 𝑦 ] 𝜓 ↔ [ 𝑥 ⇄ 𝑦 ] 𝜒 ) )

Proof

Step Hyp Ref Expression
1 ichbidv.1 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2 1 sbbidv ⊢ ( 𝜑 → ( [ 𝑎 / 𝑦 ] 𝜓 ↔ [ 𝑎 / 𝑦 ] 𝜒 ) )
3 2 sbbidv ⊢ ( 𝜑 → ( [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜓 ↔ [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜒 ) )
4 3 sbbidv ⊢ ( 𝜑 → ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜓 ↔ [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜒 ) )
5 4 1 bibi12d ⊢ ( 𝜑 → ( ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜓 ↔ 𝜓 ) ↔ ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜒 ↔ 𝜒 ) ) )
6 5 albidv ⊢ ( 𝜑 → ( ∀ 𝑦 ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜓 ↔ 𝜓 ) ↔ ∀ 𝑦 ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜒 ↔ 𝜒 ) ) )
7 6 albidv ⊢ ( 𝜑 → ( ∀ 𝑥 ∀ 𝑦 ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜓 ↔ 𝜓 ) ↔ ∀ 𝑥 ∀ 𝑦 ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜒 ↔ 𝜒 ) ) )
8 df-ich ⊢ ( [ 𝑥 ⇄ 𝑦 ] 𝜓 ↔ ∀ 𝑥 ∀ 𝑦 ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜓 ↔ 𝜓 ) )
9 df-ich ⊢ ( [ 𝑥 ⇄ 𝑦 ] 𝜒 ↔ ∀ 𝑥 ∀ 𝑦 ( [ 𝑥 / 𝑎 ] [ 𝑦 / 𝑥 ] [ 𝑎 / 𝑦 ] 𝜒 ↔ 𝜒 ) )
10 7 8 9 3bitr4g ⊢ ( 𝜑 → ( [ 𝑥 ⇄ 𝑦 ] 𝜓 ↔ [ 𝑥 ⇄ 𝑦 ] 𝜒 ) )