Main derivation
S01 · implies
Before
G=C_4,\quad Q=C_2
After
\begin{array}{c|cccc}+&0&1&2&3\\\hline 0&0&1&2&3\\1&1&2&3&0\\2&2&3&0&1\\3&3&0&1&2\\\end{array}\qquad\begin{array}{c|cc}+&0&1\\\hline 0&0&1\\1&1&0\\\end{array}Conditions: none
Rule: calc · Uses: G1
Leaf basis: calc
The two full tables contain 20 entries. Integer associativity descends; both identities are 0, with full inverse lists 0,3,2,1 and 0,1, valid on both sides.
S02 · implies
Before
f:G\to Q
After
f(0)=0,f(1)=1,f(2)=0,f(3)=1
Conditions: none
Rule: calc · Uses: G1, S01
Leaf basis: calc
Each of the four source elements has one legal target value, making a total map.
S03 · implies
Before
f((a+b)\bmod4)=((a+b)\bmod4)\bmod2
After
\begin{array}{c|cccc}f(a+b)&0&1&2&3\\\hline 0&0&1&0&1\\1&1&0&1&0\\2&0&1&0&1\\3&1&0&1&0\\\end{array}\quad=f(a)+f(b)\ (\bmod2)Conditions: none
Rule: calc · Uses: G1, S01, S02
Leaf basis: calc
All 16 pairs have the displayed value on both sides; reducing modulo 4 preserves parity, so every pair respects the operation.
S04 · implies
Before
K=f^{-1}(\{0_Q\})After
K=\{0,2\}Conditions: none
Rule: mko_definition · Uses: G1, S02, S03
Leaf basis: mko_definition
Exactly 0 and 2 map to target identity 0 in the complete mapping table.
S05 · implies
Before
K=\{0,2\}After
0\in K;\quad0+0=0,0+2=2,2+0=2,2+2=0;\quad-0=0,-2=2\ (\bmod4)
Conditions: none
Rule: calc · Uses: G1, S01, S04
Leaf basis: calc
All four sums and both inverses stay in K, directly qualifying the subgroup.
S06 · implies
Before
g=0,1,2,3,\quad k=0,2
After
\begin{array}{c|cc}g+k-g&0&2\\\hline0&0&2\\1&0&2\\2&0&2\\3&0&2\end{array}Conditions: none
Rule: calc · Uses: G1, S01, S03, S05
Leaf basis: calc
All eight conjugation entries equal k, directly proving normality rather than skipping it because of the name kernel.
S07 · equivalent
Before
f^{-1}(0)=\{0,2\},\quad f^{-1}(1)=\{1,3\}After
f(a)=f(b)\iff b-a\in K;\quad f^{-1}(0)=0+K,\ f^{-1}(1)=1+KConditions: none
Rule: calc · Uses: G1, S01, S04, S06
Leaf basis: calc
The two fibers cover G; equal parity is exactly an even residue difference, directly proving the fiber/coset equivalence.
S08 · implies
Before
0\ne2,\quad f(0)=f(2)=0
After
f\text{ not injective},\quad K\ne\{0\}Conditions: none
Rule: calc · Uses: G1, S02, S04, S07
Leaf basis: calc
The explicit collision agrees with a nontrivial kernel and refutes injectivity.
S09 · implies
Before
f(a)=f(b)\iff b-a\in K
After
K=\{0\}\iff f\text{ injective}Conditions: none
Rule: mko_definition · Uses: G1, S03, S04, S07, S08
Leaf basis: mko_definition
If K={0}, equal images force a=b; if f is injective, f(k)=f(0)=0 forces k=0. Both directions are explicit.