Biconditional Introduction: Prove Both Directions with Two Audited Subproofs

A biconditional needs two matching directions

To prove A↔B by biconditional introduction, derive B under a temporary A assumption and derive A under a separate temporary B assumption. Cite both complete ranges at their shared outer level. A single conditional direction establishes A→B, not A↔B.

This guide follows the ↔I rule of forall x: Calgary. Each opening assumption and opposite endpoint must match the two whole sides of the desired biconditional. Both blocks can use the same outer premises, but neither direction may import an individual line from its closed sibling.

Formal background: [1].

Our constructed no-premise example proves (P→Q)↔((P∨Q)→Q). One direction uses proof by cases; the other introduces a disjunction. A complete fourteen-line derivation and four-row truth check show both the formal structure and the semantic result.

Name the two whole sides before planning subproofs

A = P → Q
B = (P ∨ Q) → Q
Goal = A ↔ B

Direction 1: assume A; derive B.
Direction 2: assume B; derive A.

The A and B names are planning abbreviations for the whole displayed formulas, not new sentence letters inserted into the proof. In the first direction, B is a conditional, so its derivation needs an inner assumption P∨Q. In the second, A is a conditional, so its derivation needs an inner P assumption.

This nesting is why a two-direction outline is not enough by itself. The inner conditionals and case blocks must be completed before each direction has an endpoint at its own level.

Follow all fourteen lines with the open-assumption record

On a narrow screen, scroll inside this table. Keyboard users can focus the region and use the arrow keys.

Download the full proof, six block ranges and semantic check

Under P→Q, the inner P∨Q assumption leads to Q in both cases. The P case uses →E; the Q case merely reiterates its own assumption. ∨E puts Q back at the inner conditional’s level, so →I can export (P∨Q)→Q at line 8.

The second direction opens a sibling block at line 9. Assume P inside it, introduce P∨Q at line 11, then use the assumed conditional to get Q. Closing lines 10–12 yields P→Q at line 13. Line 14 now has the two required opposite endpoints.

Every assumption is discharged by the end, and the proof began with no premises. That makes the result a theorem of the stated package. This is different from a biconditional proved only under some undischarged factual premise.

Keep an explicit endpoint in the trivial Q case

Line 5 assumes Q and line 6 reiterates it. This makes the second case a clear Q-to-Q subproof with range 5–6. The case does not need a new premise or a special claim about Q: its own temporary assumption supplies the desired case endpoint.

Do not take line 4 from the closed P case to justify line 6 as an ordinary citation. The two cases are siblings. Although Q is the shared endpoint, the second case must establish it from its own accessible context. Reiteration from line 5 does exactly that.

Likewise, line 9 is a fresh assumption of B for the second direction, not a reuse of line 8 as an unconditional result. Line 8 still belongs to the closed first-direction block, where it depends on the temporary A assumption. The whole range 1–8 is usable for ↔I; its internal endpoint is not freely exportable.

Cross-check the equivalence on all four valuations

On a narrow screen, scroll inside this table. Keyboard users can focus the region and use the arrow keys.

Both sides are false at P=T,Q=F and true on the other three rows. A biconditional is true when its sides agree, including when both are false. Therefore the final column is all T.

This table verifies the claimed equivalence, but cannot certify every line of a proposed proof. A derivation containing an illegal closed-line reference might still print a semantically valid final formula. Audit both the formula’s truth behavior and the actual rule citations.

Reject a missing reverse direction and mismatched endpoint

A proof from P to P∨Q gives P→(P∨Q), but not P↔(P∨Q). At P=F,Q=T, P∨Q is true and P is false, so the reverse implication fails. This small countervaluation explains why one completed direction is insufficient.

For a target A↔B, a second block from B to some different C does not supply B-to-A. The endpoint must be A exactly. If you have a justified equivalence allowing a further conversion, add its licensed steps inside the block; do not silently treat two merely similar formulas as the same.

Before the final ↔I step, check both ranges, both endpoint levels, exact opposite formulas, and remaining outer premises. Afterward, use the exported biconditional at the outer level. Closed sibling lines stay closed.

Try six two-direction proof checks

Use A=P→Q and B=(P∨Q)→Q for the displayed theorem.