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The other cases are standard or similar and easier. (b) Guarded induction. We treat the cases for :P and P1 kA P2. Case :P. Assume P 2 P.
Suppose P = :P0. Then it holds that P admits a single d -transition, viz. P d P0. Thus we have P fj j P d P0 jg = = [ P0 7! ](P0) = P(P0). Suppose P = :P00 for some P00 , P. Then we have P fj j P d P0 jg = 0 = [ P00 7! ](P0) = P(P0). Case P1 kA P2. Assume P1 2 P1, P2 2 P2 and P1 kA P2 P. It holds that P = (P1 kA XP2) + (XP1 kA P2). We calculate P fj j P1 kA P2 d P0 jg = P fj j P1 d P01; P0 = P01 kA P2 jg + P fj j P2 d P02; P0 = P1 kA P02 jg (by analysis of d) = ( if P0 = P01 kA P2 then P fj j P1 d P01 jg else 0 end ) + ( if P0 = P1 kA P02 then P fj j P2 d P02 jg else 0 end ) = ( if P0 = P01 kA P2 then P1(P01) else 0 end ) + ( if P0 = P1 kA P02 then P2(P02) else 0 end ) (by induction hypothesis for P1 and P2) = ( P1 kA XP2 )(P0) + ( XP1 kA P2 )(P0) (by definition of kA, XP1,XP2 and + on FS(PIML; R>0 )) = P(P0)