3 Merge commits $L$ and $R$ using merge base $M$:
5 C \hasparents \{ L, R \}
7 \patchof{C} = \patchof{L}
11 We will occasionally use $X,Y$ s.t. $\{X,Y\} = \{L,R\}$.
13 This can also be used for dependency re-insertion, by setting
14 $L \in \pn$, $R \in \pry$, $M = \baseof{R}$.
16 \subsection{Conditions}
17 \[ \eqn{ Ingredients }{
23 R \in \py : & \baseof{R} \ge \baseof{L}
24 \land [\baseof{L} = M \lor \baseof{L} = \baseof{M}] \\
25 R \in \pn : & M = \baseof{L} \\
26 \text{otherwise} : & \false
29 \[ \eqn{ Merge Acyclic }{
34 \[ \eqn{ Removal Merge Ends }{
35 X \not\haspatch \p \land
39 \pendsof{Y}{\py} = \pendsof{M}{\py}
41 \[ \eqn{ Addition Merge Ends }{
42 X \not\haspatch \p \land
43 M \nothaspatch \p \land
46 \bigforall_{E \in \pendsof{X}{\py}} E \le Y
49 \[ \eqn{ Foreign Merges }{
50 \patchof{L} = \bot \implies \patchof{R} = \bot
53 \subsection{Non-Topbloke merges}
55 We require both $\patchof{L} = \bot$ and $\patchof{R} = \bot$
56 (Foreign Merges, above).
57 I.e. not only is it forbidden to merge into a Topbloke-controlled
58 branch without Topbloke's assistance, it is also forbidden to
59 merge any Topbloke-controlled branch into any plain git branch.
61 Given those conditions, Tip Merge and Merge Acyclic do not apply.
62 And by Foreign Contents for (wlog) Y, $\forall_{\p, D \in \py} D \not\le Y$
63 so then by No Replay $D \not\isin Y$
64 so $\neg [ Y \haspatch \p ]$ so neither
65 Merge Ends condition applies.
67 So a plain git merge of non-Topbloke branches meets the conditions and
68 is therefore consistent with our model.
70 \subsection{No Replay}
72 By definition of $\merge$,
73 $D \isin C \implies D \isin L \lor D \isin R \lor D = C$.
75 Ingredients Prevent Replay applies. $\qed$
77 \subsection{Unique Base}
79 Need to consider only $C \in \py$, ie $L \in \py$,
80 and calculate $\pendsof{C}{\pn}$. So we will consider some
81 putative ancestor $A \in \pn$ and see whether $A \le C$.
83 By Exact Ancestors for C, $A \le C \equiv A \le L \lor A \le R \lor A = C$.
84 But $C \in \py$ and $A \in \pn$ so $A \neq C$.
85 Thus $A \le C \equiv A \le L \lor A \le R$.
87 By Unique Base of L and Transitive Ancestors,
88 $A \le L \equiv A \le \baseof{L}$.
90 \subsubsection{For $R \in \py$:}
92 By Unique Base of $R$ and Transitive Ancestors,
93 $A \le R \equiv A \le \baseof{R}$.
95 But by Tip Merge condition on $\baseof{R}$,
96 $A \le \baseof{L} \implies A \le \baseof{R}$, so
97 $A \le \baseof{R} \lor A \le \baseof{L} \equiv A \le \baseof{R}$.
98 Thus $A \le C \equiv A \le \baseof{R}$.
99 That is, $\baseof{C} = \baseof{R}$.
101 \subsubsection{For $R \in \pn$:}
103 By Tip Merge condition and since $M \le R$,
104 $A \le \baseof{L} \implies A \le R$, so
105 $A \le R \lor A \le \baseof{L} \equiv A \le R$.
106 Thus $A \le C \equiv A \le R$.
107 That is, $\baseof{C} = R$.
111 \subsection{Coherence and Patch Inclusion}
113 Need to determine $C \haspatch \p$ based on $L,M,R \haspatch \p$.
114 This involves considering $D \in \py$.
116 \subsubsection{For $L \nothaspatch \p, R \nothaspatch \p$:}
117 $D \not\isin L \land D \not\isin R$. $C \not\in \py$ (otherwise $L
118 \in \py$ ie $L \haspatch \p$ by Tip Self Inpatch for $L$). So $D \neq C$.
119 Applying $\merge$ gives $D \not\isin C$ i.e. $C \nothaspatch \p$.
121 \subsubsection{For $L \haspatch \p, R \haspatch \p$:}
122 $D \isin L \equiv D \le L$ and $D \isin R \equiv D \le R$.
123 (Likewise $D \isin X \equiv D \le X$ and $D \isin Y \equiv D \le Y$.)
125 Consider $D = C$: $D \isin C$, $D \le C$, OK for $C \haspatch \p$.
127 For $D \neq C$: $D \le C \equiv D \le L \lor D \le R
128 \equiv D \isin L \lor D \isin R$.
129 (Likewise $D \le C \equiv D \le X \lor D \le Y$.)
131 Consider $D \neq C, D \isin X \land D \isin Y$:
132 By $\merge$, $D \isin C$. Also $D \le X$
133 so $D \le C$. OK for $C \haspatch \p$.
135 Consider $D \neq C, D \not\isin X \land D \not\isin Y$:
136 By $\merge$, $D \not\isin C$.
137 And $D \not\le X \land D \not\le Y$ so $D \not\le C$.
138 OK for $C \haspatch \p$.
140 Remaining case, wlog, is $D \not\isin X \land D \isin Y$.
141 $D \not\le X$ so $D \not\le M$ so $D \not\isin M$.
142 Thus by $\merge$, $D \isin C$. And $D \le Y$ so $D \le C$.
143 OK for $C \haspatch \p$.
145 So indeed $L \haspatch \p \land R \haspatch \p \implies C \haspatch \p$.
147 \subsubsection{For (wlog) $X \not\haspatch \p, Y \haspatch \p$:}
149 $M \haspatch \p \implies C \nothaspatch \p$.
150 $M \nothaspatch \p \implies C \haspatch \p$.
154 One of the Merge Ends conditions applies.
155 Recall that we are considering $D \in \py$.
156 $D \isin Y \equiv D \le Y$. $D \not\isin X$.
157 We will show for each of
158 various cases that $D \isin C \equiv M \nothaspatch \p \land D \le C$
159 (which suffices by definition of $\haspatch$ and $\nothaspatch$).
161 Consider $D = C$: Thus $C \in \py, L \in \py$. By Tip Contents
162 for $L$, $L \isin L$ so $\neg [ L \nothaspatch \p ]$.
163 Therefore we must have $L=Y$, $R=X$.
164 By Tip Merge $M = \baseof{L}$ so $M \in \pn$ so
165 by Base Acyclic $M \nothaspatch \p$. By $\merge$, $D \isin C$,
166 and $D \le C$, consistent with $C \haspatch \p$. OK.
168 Consider $D \neq C, M \nothaspatch \p, D \isin Y$:
169 $D \le Y$ so $D \le C$.
170 $D \not\isin M$ so by $\merge$, $D \isin C$. OK.
172 Consider $D \neq C, M \nothaspatch \p, D \not\isin Y$:
173 $D \not\le Y$. If $D \le X$ then
174 $D \in \pancsof{X}{\py}$, so by Addition Merge Ends and
175 Transitive Ancestors $D \le Y$ --- a contradiction, so $D \not\le X$.
176 Thus $D \not\le C$. By $\merge$, $D \not\isin C$. OK.
178 Consider $D \neq C, M \haspatch \p, D \isin Y$:
179 $D \le Y$ so $D \in \pancsof{Y}{\py}$ so by Removal Merge Ends
180 and Transitive Ancestors $D \in \pancsof{M}{\py}$ so $D \le M$.
181 Thus $D \isin M$. By $\merge$, $D \not\isin C$. OK.
183 Consider $D \neq C, M \haspatch \p, D \not\isin Y$:
184 By $\merge$, $D \not\isin C$. OK.
188 \subsection{Base Acyclic}
190 This applies when $C \in \pn$.
191 $C \in \pn$ when $L \in \pn$ so by Merge Acyclic, $R \nothaspatch \p$.
193 Consider some $D \in \py$.
195 By Base Acyclic of $L$, $D \not\isin L$. By the above, $D \not\isin
196 R$. And $D \neq C$. So $D \not\isin C$.
200 \subsection{Tip Contents}
202 We need worry only about $C \in \py$.
203 And $\patchof{C} = \patchof{L}$
204 so $L \in \py$ so $L \haspatch \p$. We will use the Unique Base
205 of $C$, and its Coherence and Patch Inclusion, as just proved.
207 Firstly we show $C \haspatch \p$: If $R \in \py$, then $R \haspatch
208 \p$ and by Coherence/Inclusion $C \haspatch \p$ . If $R \not\in \py$
209 then by Tip Merge $M = \baseof{L}$ so by Base Acyclic and definition
210 of $\nothaspatch$, $M \nothaspatch \p$. So by Coherence/Inclusion $C
211 \haspatch \p$ (whether $R \haspatch \p$ or $\nothaspatch$).
213 We will consider an arbitrary commit $D$
214 and prove the Exclusive Tip Contents form.
216 \subsubsection{For $D \in \py$:}
217 $C \haspatch \p$ so by definition of $\haspatch$, $D \isin C \equiv D
220 \subsubsection{For $D \not\in \py, R \not\in \py$:}
222 $D \neq C$. By Tip Contents of $L$,
223 $D \isin L \equiv D \isin \baseof{L}$, so by Tip Merge condition,
224 $D \isin L \equiv D \isin M$. So by $\merge$, $D \isin
225 C \equiv D \isin R$. And $R = \baseof{C}$ by Unique Base of $C$.
226 Thus $D \isin C \equiv D \isin \baseof{C}$. OK.
228 \subsubsection{For $D \not\in \py, R \in \py$:}
233 $D \isin L \equiv D \isin \baseof{L}$ and
234 $D \isin R \equiv D \isin \baseof{R}$.
236 Apply Tip Merge condition.
237 If $\baseof{L} = M$, trivially $D \isin M \equiv D \isin \baseof{L}.$
238 Whereas if $\baseof{L} = \baseof{M}$, by definition of $\base$,
239 $\patchof{M} = \patchof{L} = \py$, so by Tip Contents of $M$,
240 $D \isin M \equiv D \isin \baseof{M} \equiv D \isin \baseof{L}$.
242 So $D \isin M \equiv D \isin L$ so by $\merge$,
243 $D \isin C \equiv D \isin R$. But from Unique Base,
244 $\baseof{C} = \baseof{R}$.
245 Therefore $D \isin C \equiv D \isin \baseof{C}$. OK.
249 \subsection{Foreign Inclusion}
251 Consider some $D$ s.t. $\patchof{D} = \bot$.
252 By Foreign Inclusion of $L, M, R$:
253 $D \isin L \equiv D \le L$;
254 $D \isin M \equiv D \le M$;
255 $D \isin R \equiv D \le R$.
257 \subsubsection{For $D = C$:}
259 $D \isin C$ and $D \le C$. OK.
261 \subsubsection{For $D \neq C, D \isin M$:}
263 Thus $D \le M$ so $D \le L$ and $D \le R$ so $D \isin L$ and $D \isin
264 R$. So by $\merge$, $D \isin C$. And $D \le C$. OK.
266 \subsubsection{For $D \neq C, D \not\isin M, D \isin X$:}
268 By $\merge$, $D \isin C$.
269 And $D \isin X$ means $D \le X$ so $D \le C$.
272 \subsubsection{For $D \neq C, D \not\isin M, D \not\isin L, D \not\isin R$:}
274 By $\merge$, $D \not\isin C$.
275 And $D \not\le L, D \not\le R$ so $D \not\le C$.
280 \subsection{Foreign Contents}
282 Only relevant if $\patchof{L} = \bot$, in which case
283 $\patchof{C} = \bot$ and by Foreign Merges $\patchof{R} = \bot$,
284 so Totally Foreign Contents applies. $\qed$