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Mirrors > Home > ILE Home > Th. List > hbbi | GIF version |
Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑 ↔ 𝜓). (Contributed by NM, 5-Aug-1993.) |
Ref | Expression |
---|---|
hb.1 | ⊢ (𝜑 → ∀𝑥𝜑) |
hb.2 | ⊢ (𝜓 → ∀𝑥𝜓) |
Ref | Expression |
---|---|
hbbi | ⊢ ((𝜑 ↔ 𝜓) → ∀𝑥(𝜑 ↔ 𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | dfbi2 388 | . 2 ⊢ ((𝜑 ↔ 𝜓) ↔ ((𝜑 → 𝜓) ∧ (𝜓 → 𝜑))) | |
2 | hb.1 | . . . 4 ⊢ (𝜑 → ∀𝑥𝜑) | |
3 | hb.2 | . . . 4 ⊢ (𝜓 → ∀𝑥𝜓) | |
4 | 2, 3 | hbim 1545 | . . 3 ⊢ ((𝜑 → 𝜓) → ∀𝑥(𝜑 → 𝜓)) |
5 | 3, 2 | hbim 1545 | . . 3 ⊢ ((𝜓 → 𝜑) → ∀𝑥(𝜓 → 𝜑)) |
6 | 4, 5 | hban 1547 | . 2 ⊢ (((𝜑 → 𝜓) ∧ (𝜓 → 𝜑)) → ∀𝑥((𝜑 → 𝜓) ∧ (𝜓 → 𝜑))) |
7 | 1, 6 | hbxfrbi 1472 | 1 ⊢ ((𝜑 ↔ 𝜓) → ∀𝑥(𝜑 ↔ 𝜓)) |
Colors of variables: wff set class |
Syntax hints: → wi 4 ∧ wa 104 ↔ wb 105 ∀wal 1351 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-ia1 106 ax-ia2 107 ax-ia3 108 ax-5 1447 ax-gen 1449 ax-4 1510 ax-i5r 1535 |
This theorem depends on definitions: df-bi 117 |
This theorem is referenced by: euf 2031 sb8euh 2049 |
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