Main derivation
S01 · definition
Before
\text{a candidate sentence}After
\text{proposition}\iff\text{well-formed declarative sentence with no unassigned free variable}Conditions: none
Rule: mko_prop · Uses: G1
Leaf basis: mko_prop + prop_test
By the MKO, this guide uses an introductory classical context: a proposition is a statement that can be asserted or denied, and in this context it is true or false; but being a proposition does not mean being proved true. Whether a candidate is a proposition is decided by its form (whether it is a closed declarative sentence, prop_test), not by its truth. For each candidate, first decide whether it is a proposition, then separately decide its truth.
S02 · implies
Before
s_1:\ 7\ \text{is prime}After
s_1\ \text{is a proposition}Conditions: none
Rule: prop_test · Uses: S01, G2
Leaf basis: prop_test
s1 is a declarative sentence asserting a property of the definite integer 7, with no unassigned variable, so by prop_test it is a proposition. At this point its truth has not been decided, let alone proved; its status as a proposition is settled already here.
S03 · equivalent
Before
7\ \text{is prime}After
7\ge 2\ \land\ \neg\exists d\in\mathbb{Z}\,(2\le d\le 6\ \land\ d\mid 7)Conditions: none
Rule: prime_def · Uses: S02, G3
Leaf basis: prime_def + def_unfold
Unfold s1 by the definition of prime in G3 (p=7, p−1=6). The first conjunct, 7≥2, holds; what remains is that no integer d with 2≤d≤6 satisfies d∣7.
S04 · implies
Before
d\in\{2,3,4,5,6\}After
d\nmid 7\ \text{for each such}\ dConditions: none
Rule: L_rem · Uses: S03, int_no_gap, division_algorithm
Leaf basis: division_algorithm + divides_def + int_no_gap + numeric_evaluation
The integers with 2≤d≤6 are exactly 2, 3, 4, 5, 6 (there is no integer between consecutive integers, int_no_gap), a finite range listed completely. For each d write the division with remainder: 7=2·3+1, 7=3·2+1, 7=4·1+3, 7=5·1+2, 7=6·1+1, with remainders 1, 1, 3, 2, 1, each satisfying 0<r<d. By Lemma L_rem none of these five values of d divides 7.
S05 · implies
Before
7\ge 2,\ \text{and no}\ d\ \text{with}\ 2\le d\le 6\ \text{divides}\ 7After
s_1\ \text{is true, and now proved}Conditions: none
Rule: and_intro_elim · Uses: S03, S04
Leaf basis: and_intro_elim + quantifier_negation
Both conjuncts now hold: 7≥2, and S04 shows d∤7 for every d in the range, so no such d exists. By the equivalence S03, s1 is true, and it now has a proof.
S06 · implies
Before
s_2:\ 4\ \text{is prime}After
s_2\ \text{is a proposition, and}\ \neg s_2\ \text{holds}Conditions: none
Rule: prime_def · Uses: G2, G3, prop_test
Leaf basis: prime_def + divides_def + exists_intro + numeric_evaluation
s2 is also a closed declarative sentence, so by prop_test it is a proposition. By the definition, 4 is prime only if no integer d with 2≤d≤3 divides 4; but d=2 satisfies 2≤2≤3 and 4=2·2, so 2∣4. Hence s2 is false, and what is proved is its negation ¬s2. s2 is a false proposition: being a proposition and being true are different things.
S07 · implies
Before
s_3:\ \text{Is 7 prime?}After
s_3\ \text{is not a proposition}Conditions: none
Rule: prop_test · Uses: G2, prop_test
Leaf basis: prop_test
s3 is a question; it asserts nothing and is neither true nor false, so by prop_test it is not a proposition. What it asks about corresponds to the proposition s1; answering 'yes' rests on the proof of s1 in S05, not on the question itself being true.
S08 · implies
Before
s_4:\ \text{Compute}\ 3+4.After
s_4\ \text{is not a proposition}Conditions: none
Rule: prop_test · Uses: G2, prop_test
Leaf basis: prop_test + numeric_evaluation
s4 is a command asking for a computation; it asserts nothing, so by prop_test it is not a proposition. By contrast, the declarative sentence '3+4=7' is a true proposition (N14).
S09 · implies
Before
s_5:\ x>0
After
s_5\ \text{is an open sentence, not a proposition}Conditions: none
Rule: prop_test · Uses: G2, S01, prop_test
Leaf basis: prop_test
s5 is declarative, but x is an unassigned free variable, so by prop_test it is an open sentence: its truth depends on the value of x (S10 shows it true at x=2 and false at x=−1), so in a closed context it is not itself a proposition. To get a proposition it must be closed by substitution or quantification (the MKO's third paragraph).
S10 · implies
Before
x>0\ \text{with}\ x:=2\ \text{or}\ x:=-1After
2>0\ (\mathrm{true}),\quad -1>0\ (\mathrm{false})Conditions: none
Rule: order_basics · Uses: S09
Leaf basis: order_basics + two_nonzero
Substitution is one way to close it. x:=2 gives the proposition '2>0', which is true (0<1<2); x:=−1 gives '−1>0', which is false: from 0<1, adding −1 gives −1<0, and trichotomy rules out −1>0. Different substitutions into one open sentence give two different propositions with different truth values.
S11 · implies
Before
\forall x\in\mathbb{Z},\ x>0After
\text{false: counterexample}\ x=0Conditions: none
Rule: quantifier_negation · Uses: S09
Leaf basis: quantifier_negation + exists_intro + order_basics
Universal quantification is another way to close it: '∀x∈ℤ, x>0' has no free variable and is a proposition. Take x=0: 0=0, so by trichotomy 0>0 fails; hence ∃x∈ℤ, ¬(x>0), and by negation of quantifiers ¬∀x∈ℤ, x>0, so the original universally quantified proposition is false. However many positive cases such as 1>0, 2>0, 3>0 are listed, they cannot prove this universal statement, while one counterexample refutes it.
S12 · implies
Before
\exists x\in\mathbb{Z},\ x>0After
\text{true: witness}\ x=1Conditions: none
Rule: exists_intro · Uses: S09
Leaf basis: exists_intro + two_nonzero
Existential quantification: '∃x∈ℤ, x>0' is a proposition. Take x=1; 1>0 holds (0<1), so by ∃-introduction the proposition is true.
S13 · implies
Before
\text{verdicts of S02 to S12}After
\text{propositions: }s_1,s_2,\text{closures of }s_5;\ \text{not propositions: }s_3,s_4,s_5Conditions: none
Rule: and_intro_elim · Uses: S02, S05, S06, S07, S08, S09, S10, S11, S12
Leaf basis: and_intro_elim
Propositions: s1 (true, proved in S05), s2 (false, its negation proved in S06) and the four closed versions of s5 (2>0 true, −1>0 false, the ∀ version false, the ∃ version true). Not propositions: s3 (a question), s4 (a command) and s5 itself (an open sentence). For every candidate, proposition status was settled before truth.