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Metamath Proof Explorer

Theorem 2sbbid

Description: Deduction doubly substituting both sides of a biconditional. (Contributed by AV, 30-Jul-2023)

Ref Expression
Hypotheses sbbid.1 ⊢ Ⅎ 𝑥 𝜑
sbbid.2 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
2sbbid.1 ⊢ Ⅎ 𝑦 𝜑
Assertion 2sbbid ( 𝜑 → ( [ 𝑡 / 𝑥 ] [ 𝑢 / 𝑦 ] 𝜓 ↔ [ 𝑡 / 𝑥 ] [ 𝑢 / 𝑦 ] 𝜒 ) )

Proof

Step Hyp Ref Expression
1 sbbid.1 ⊢ Ⅎ 𝑥 𝜑
2 sbbid.2 ⊢ ( 𝜑 → ( 𝜓 ↔ 𝜒 ) )
3 2sbbid.1 ⊢ Ⅎ 𝑦 𝜑
4 3 2 sbbid ⊢ ( 𝜑 → ( [ 𝑢 / 𝑦 ] 𝜓 ↔ [ 𝑢 / 𝑦 ] 𝜒 ) )
5 1 4 sbbid ⊢ ( 𝜑 → ( [ 𝑡 / 𝑥 ] [ 𝑢 / 𝑦 ] 𝜓 ↔ [ 𝑡 / 𝑥 ] [ 𝑢 / 𝑦 ] 𝜒 ) )