Classical modus ponens needs P as its first premise. You can swap in the weaker premise P∨Q and still reach Q. Drawn as sixteen-valued squares, this version gives back exactly Q, with nothing left over and nothing left free.
P∨Q is the weakest first premise that works: every premise that works fits inside it. Classical modus ponens, with P, is one of those premises.
Each value is a 2×2 square in the Gates of gates colours. A quarter is coloured when the value is TRUE on that row. P is value 6, Q is value 3.
Compare classical modus ponens. There the premises combine to P∧Q, which fits inside Q but is smaller. It has the TT quarter and misses the FT quarter:
On the rows where P is true, P→Q reduces to Q. On the rows where P is false, P∨Q reduces to Q. Together they cover all four rows, so Q is pinned everywhere:
Classical modus ponens has only the first half. Its premise P is false on the FT and FF rows, and there it says nothing about Q.
Put any square X in (X ∧ (P→Q)) → Q. It is a tautology exactly when X leaves the FF quarter empty, which means X fits inside P∨Q. The eight that work form a cube, with P∨Q at the top. Under each is what X ∧ (P→Q) becomes.
The other eight all fill the FF quarter, outlined in red. On that row P and Q are both false, P→Q is true, and so the premises hold while Q fails:
Now let P and Q be any of the sixteen values. Pick them below. The identity (P∨Q) ∧ (P→Q) = Q holds every time.
| classical: P, P→Q | this version: P∨Q, P→Q |
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Resolution. P→Q is ¬P∨Q, so the argument is “P∨Q and ¬P∨Q, therefore Q”. This is the resolution rule, which cancels P against ¬P.
Proof by cases. Either P or Q holds. If P, then Q by P→Q. If Q, then Q. Both cases end at Q.
Modus ponens as a special case. P fits inside P∨Q, so any argument that starts from P also starts from P∨Q. Starting from P gives the smaller result P∧Q. Starting from P∨Q gives all of Q.