Main derivation
S01 · definition
Before
H\subseteq G,\ f:G\to K
After
H\neq\varnothing,\ HH\subseteq H,\ e\in H,\ H^{-1}\subseteq H;\ \forall a,b:\ f(a*b)=f(a)\oplus f(b)Conditions: none
Rule: mko_subhom_def · Uses: mko_subhom_def, G1, G2
Leaf basis: mko_subhom_def + group_axioms
By the MKO a subgroup needs four checks: nonempty, closed, contains the identity, closed under inverses; a homomorphism must satisfy f(a*b)=f(a)⊕f(b) for every input pair. The MKO also warns that a homomorphism is not automatically injective.
S02 · implies
Before
H=\{0,2\}After
H\neq\varnothing,\ H\subseteq G
Conditions: none
Rule: subset_def · Uses: G1, G4
Leaf basis: subset_def
H contains 0, so it is nonempty; both of its elements 0 and 2 lie in G, so H⊆G.
S03 · implies
Before
a,b\in H
After
0*0=0,\ 0*2=2,\ 2*0=2,\ 2*2=0\ \in H
Conditions: none
Rule: finite_exhaustion · Uses: G1, S02
Leaf basis: finite_exhaustion + numeric_evaluation
H has 2²=4 ordered pairs; reading G's table: 0*0=0, 0*2=2, 2*0=2 and 2*2=(2+2) mod 4=0. All four outputs lie in H, so H is closed.
S04 · implies
Before
e_G=0
After
e_G\in H
Conditions: none
Rule: set_builder · Uses: G1, S02
Leaf basis: group_axioms + set_builder
The identity of G is 0 (row 0 and column 0 of the table are both 0,1,2,3), and 0∈H.
S05 · implies
Before
a\in H
After
0^{-1}=0\in H,\ 2^{-1}=2\in HConditions: none
Rule: finite_exhaustion · Uses: G1, S04
Leaf basis: finite_exhaustion + group_axioms
0*0=0, so the inverse of 0 is 0; 2*2=0 on both sides, so the two-sided inverse of 2 is 2. Both inverses lie in H.
S06 · implies
Before
\text{nonempty, closed, identity, inverses}After
H\ \text{is a subgroup of } GConditions: none
Rule: mko_subhom_def · Uses: S02, S03, S04, S05
Leaf basis: mko_subhom_def + and_intro_elim
The four conditions come from S02 to S05, so by the MKO's definition H is a subgroup of G.
S07 · implies
Before
H'=\{0,1\},\ 1*1After
1*1=2\notin H'\ \Rightarrow\ H'\ \text{not a subgroup}Conditions: none
Rule: mko_subhom_def · Uses: G1, G4
Leaf basis: mko_subhom_def + quantifier_negation + numeric_evaluation
H' is nonempty and contains the identity 0, but 1*1=(1+1) mod 4=2 is not in H', so closure fails and H' is not a subgroup. Moreover the inverse of 1 is 3, which is not in H' either.
S08 · implies
Before
f(0)=0,\ f(1)=1,\ f(2)=0,\ f(3)=1
After
f:G\to K,\ f(a)=a\bmod 2
Conditions: none
Rule: function_def · Uses: G3, G2, mod_notation
Leaf basis: function_def + finite_exhaustion + mod_notation
The mapping table assigns exactly one value in K to each element of G, so f is a function G→K. Comparing the four cells: the remainders of 0, 1, 2, 3 on division by 2 are 0, 1, 0, 1, so f(a)=a mod 2.
S09 · implies
Before
x-y=mk,\ k\in\mathbb{Z}After
x\bmod m=y\bmod m
Conditions: m>0
Rule: L_same_rem · Uses: division_algorithm, mod_notation
Leaf basis: division_algorithm + mod_notation
The homomorphism argument needs Lemma L_same_rem: two integers that differ by an integer multiple of m leave the same remainder on division by m. The proof is in S09.1 and S09.2 and uses only existence and uniqueness in the division algorithm.
S10 · equivalent
Before
f(a*b)=((a+b)\bmod 4)\bmod 2
After
(a+b)\bmod 2
Conditions: a,b\in G
Rule: L_same_rem · Uses: S08, S09, G1, division_algorithm
Leaf basis: L_same_rem + division_algorithm + divides_def + eq_replace
Write a+b=4q+s with s=(a+b) mod 4. Then s−(a+b)=−4q=2·(−2q) is an integer multiple of 2, so by the lemma (m=2), s mod 2=(a+b) mod 2. Here a and b are arbitrary elements of G.
S11 · equivalent
Before
f(a)\oplus f(b)=(a\bmod 2+b\bmod 2)\bmod 2
After
(a+b)\bmod 2
Conditions: a,b\in G
Rule: L_same_rem · Uses: S08, S09, G2, division_algorithm
Leaf basis: L_same_rem + division_algorithm + eq_replace
Write a=2p+r₁ and b=2p'+r₂ (r₁=a mod 2, r₂=b mod 2). Then (r₁+r₂)−(a+b)=−2(p+p') is a multiple of 2, so by the lemma (r₁+r₂) mod 2=(a+b) mod 2.
S12 · implies
Before
f(a*b)=(a+b)\bmod 2=f(a)\oplus f(b)
After
\forall a,b\in G:\ f(a*b)=f(a)\oplus f(b)
Conditions: none
Rule: eq_refl_symm_trans · Uses: S10, S11
Leaf basis: eq_refl_symm_trans + forall_intro
Both sides equal (a+b) mod 2, so they are equal by transitivity. S10 and S11 hold for arbitrary a, b, so all 4²=16 input pairs are covered (MKO paragraph 2). N5 separately enumerates the 16 pairs as a cross-check. f is a homomorphism.
S13 · equivalent
Before
\ker f=\{a\in G\mid f(a)=0\}After
\{0,2\}=HConditions: none
Rule: kernel_image_def · Uses: S08, G2, kernel_image_def
Leaf basis: kernel_image_def + set_builder + finite_exhaustion
The target identity is 0. Checking each element: f(0)=0, f(1)=1, f(2)=0, f(3)=1, so ker f={0,2}, which is exactly the subgroup H of S06.
S14 · implies
Before
\operatorname{im} f,\ f(0)=f(2)After
\operatorname{im} f=\{0,1\}=K;\ f\ \text{surjective, not injective}Conditions: none
Rule: injective_surjective_def · Uses: S08, S13, kernel_image_def
Leaf basis: injective_surjective_def + kernel_image_def + finite_counting_basics
The image, all mapped values, is {0,1}=K, so f is surjective. But f(0)=f(2)=0 with 0≠2, so f is not injective; the kernel containing more than the identity reflects this. |im f|=2<4=|G|, and the MKO's third paragraph says injectivity is possible only when the image is as large as the source. f is not a bijection, hence not an isomorphism.
S15 · implies
Before
h(0*0)\ \text{vs}\ h(0)\oplus h(0)After
1\neq 0=1\oplus 1\ \Rightarrow\ h\ \text{not a homomorphism}Conditions: none
Rule: quantifier_negation · Uses: G3, G1, G2
Leaf basis: quantifier_negation + numeric_evaluation
h(0*0)=h(0)=1, but h(0)⊕h(0)=1⊕1=(1+1) mod 2=0. Since 1≠0, the homomorphism condition fails at (0,0); by quantifier negation h is not a homomorphism. One failing pair suffices to reject, whereas establishing a homomorphism requires an argument covering all 16 pairs.
S16 · implies
Before
H,\ H',\ f,\ h
After
H\le G,\ H'\not\le G,\ f\ \text{hom.},\ \ker f=H,\ \operatorname{im} f=K,\ f\ \text{not injective},\ h\ \text{not hom.}Conditions: none
Rule: and_intro_elim · Uses: S06, S07, S12, S13, S14, S15
Leaf basis: and_intro_elim
Combine all conclusions; they concern only the two specified finite groups G and K.