DISJ_CASES : (thm -> thm -> thm -> thm)
A |- t1 \/ t2 A1 u {t1} |- t A2 u {t2} |- t ------------------------------------------------------ DISJ_CASES A u A1 u A2 |- t
th = |- (m = 0) \/ (?n. m = SUC n)Using two additional theorems, each having one disjunct as a hypothesis:
th1 = (m = 0 |- (PRE m = m) = (m = 0)) th2 = (?n. m = SUC n" |- (PRE m = m) = (m = 0))a new theorem can be derived:
#DISJ_CASES th th1 th2;; |- (PRE m = m) = (m = 0)