Assessment Solutions and Scoring

Use only after saving an independent checkpoint attempt. Score each competency with the rubric. The essential steps below distinguish a full argument from an unsupported final answer; alternative valid proofs count.

A1 — Logic and scope

Negation: ∃x∈Z ∀y∈Z: y≤x or y is not even. For the original choose y=2(|x|+1), an even integer greater than x: if x≥0 it is 2x+2>x; if x<0 it is positive and hence greater than x. The swapped claim fails by choosing x=y against any proposed y. With both domains empty the original universal claim is true and the swapped existential claim false. C1 needs the conjunction's correct negation and permitted witness dependency, not just true/false.

A2 — Proof construction

If exactly one integer is odd, write the pair as 2j and 2k+1 in either order; the sum is 2(j+k)+1. Conversely, if both are even their sum is 2(j+k); if both are odd it is 2(j+k+1). Thus an odd sum excludes both equal-parity cases and forces exactly one odd. Parity is exhaustive for integers. The product claim fails at a=2,b=1. C2 needs both directions, coverage, and a counterexample whose product is even while at least one factor is odd.

A3 — Sets

If x∈(A∪B)∖C then x∉C and x belongs to A or B. In the first case x∈A∖C; in the second x∈B∖C; hence it belongs to the right side. Conversely, either right-side membership gives x∉C and x∈A∪B, hence membership on the left. The sets {1} and {1,2} show one inclusion can be strict. C3 requires the arbitrary-element argument in both directions (a correct iff chain also works).

A4 — Functions

f is onto B but not injective, since 1 and 2 collide at 11. g is bijective: its two inputs have distinct outputs covering C. Composite values are 20,21,21, so g∘f is onto C but not injective. The preimage of {11} is {1,2}. In general, for any c∈C, composite surjectivity supplies a∈A with g(f(a))=c. Then b=f(a)∈B witnesses surjectivity of g. f need not be onto B: take A={0}, B={1,2}, C={9}, f(0)=1, g(1)=g(2)=9. The composite is onto C but 2 is absent from f's image. C4 needs a typed counterexample and general proof.

A5 — Relations

Reflexive and antisymmetric; not symmetric because (0,1) lacks its reverse; not transitive because 0R1 and 1R2 but not 0R2. Antisymmetry holds because no distinct pair has both directions. R∘R=R∪{(0,2)}: diagonal middle witnesses preserve existing pairs, and the only new two-step pair is (0,2). This relation is already transitive and contains R, so it is both R+ and R* here. For the empty relation on {0}, positive closure is empty whereas reflexive-transitive closure is {(0,0)}. C5 needs all four justifications and the closure convention.

A6 — Quotients and orders

Equality of absolute values is reflexive, symmetric and transitive by equality of integers. Classes are Z0={0}, Z1={−1,1}, Z2={−2,2}. The map to x is ill-defined on Z1 or Z2; the square map is well-defined because equal absolute values have equal squares. Ordering by absolute value is independent of the representative, and yields Z0≤Z1≤Z2. Covers are (Z0,Z1),(Z1,Z2); least/minimal is Z0, greatest/maximal is Z2. C6 requires a representative-independence argument, not just a diagram. Class names Z0,Z1,Z2 here are local labels, not integer sets.

A7 — Induction and termination

For the sum, n=0 gives 1=2−1. Assume the sum through n is 2^(n+1)−1; adding 2^(n+1) gives 2^(n+2)−1. Induction establishes all n≥0. For 3a+5b, bases are 8=3+5, 9=3·3, 10=5·2. For n≥11 use the strong hypothesis at n−3≥8, then add one 3. This produces nonnegative coefficients. The sequence 1,1/2,1/4,... decreases forever among positive reals; a natural number measure or another well-founded argument is needed. C7 requires both proofs with justified starting cases and a valid termination distinction.

A8 — Structural proof and execution

Define mirror(leaf(v))=leaf(v) and mirror(node(l,r))=node(mirror(r),mirror(l)). The leaf case is immediate. At a node, applying mirror twice yields node(mirror(mirror(l)),mirror(mirror(r))). The two structural hypotheses reduce this to node(l,r). These are all constructors, so the theorem holds for every finite full binary tree under structural equality. It does not establish available call-stack space; an explicit stack with pending constructor records can avoid recursive call depth. Memory can still be exhausted. A shared leaf appearing in both child positions contributes two leaf occurrences; counting object identities instead describes a different representation property. C8 needs constructor cases, both hypotheses and the execution/model distinction.