(UNSET-WATERFALL-PARALLELISM) (ASSIGN SCRIPT-MODE T) T (SET-LD-PROMPT T STATE) T ACL2 !>>(SET-INHIBITED-SUMMARY-TYPES '(TIME STEPS)) (TIME STEPS) ACL2 !>>(SET-INHIBIT-OUTPUT-LST '(PROOF-TREE)) (PROOF-TREE) ACL2 !>>(ER-PROGN (SET-TRACE-CO (@ STANDARD-CO) STATE) (VALUE NIL)) NIL ACL2 !>>(DEFN F1 (X) (LIST (CAR (CONS X X)))) Since F1 is non-recursive, its admission is trivial. We observe that the type of F1 is described by the theorem (AND (CONSP (F1 X)) (TRUE-LISTP (F1 X))). We used primitive type reasoning. Computing the guard conjecture for F1.... The guard conjecture for F1 is trivial to prove, given primitive type reasoning. F1 is compliant with Common Lisp. Summary Form: ( DEFUN F1 ...) Rules: ((:FAKE-RUNE-FOR-TYPE-SET NIL)) F1 ACL2 !>>(DEFN G1 (X) (LIST X)) Since G1 is non-recursive, its admission is trivial. We observe that the type of G1 is described by the theorem (AND (CONSP (G1 X)) (TRUE-LISTP (G1 X))). We used primitive type reasoning. Computing the guard conjecture for G1.... The guard conjecture for G1 is trivial to prove. G1 is compliant with Common Lisp. Summary Form: ( DEFUN G1 ...) Rules: ((:FAKE-RUNE-FOR-TYPE-SET NIL)) G1 ACL2 !>>(MEMOIZE 'F1 :INVOKE 'G1) ACL2 Error in ( TABLE MEMOIZE-TABLE ...): Illegal attempt to set memoize- table: The following event must be admitted (possibly with differing name or macro) before memoizing function F1 with :INVOKE value G1. See :DOC memoize. (DEFTHM |F1-is-G1| (EQUAL (F1 X) (G1 X)) :RULE-CLASSES NIL) ACL2 Error [Failure] in ( TABLE MEMOIZE-TABLE ...): See :DOC failure. ACL2 !>>(DEFTHM |F1-is-G1| (EQUAL (F1 X) (G1 X)) :RULE-CLASSES NIL) Goal' Q.E.D. Summary Form: ( DEFTHM |F1-is-G1| ...) Rules: ((:DEFINITION F1) (:DEFINITION G1) (:FAKE-RUNE-FOR-TYPE-SET NIL) (:REWRITE CAR-CONS)) |F1-is-G1| ACL2 !>>(MEMOIZE 'F1 :INVOKE 'G1) F1 ACL2 !>>(TRACE$ G1) ((G1)) ACL2 !>>(F1 3) 1> (G1 3) <1 (G1 (3)) (3) ACL2 !>>(DEFUN FIB (N) (DECLARE (XARGS :GUARD (NATP N))) (IF (ZP N) 0 (IF (EQL N 1) 1 (+ (FIB (- N 1)) (FIB (- N 2)))))) For the admission of FIB we will use the relation O< (which is known to be well-founded on the domain recognized by O-P) and the measure (ACL2-COUNT N). The non-trivial part of the measure conjecture is Goal (IMPLIES (AND (NOT (ZP N)) (NOT (EQL N 1))) (O< (ACL2-COUNT (+ -2 N)) (ACL2-COUNT N))). Goal' Splitter note (see :DOC splitter) for Goal' (2 subgoals). if-intro: ((:DEFINITION ACL2-COUNT) (:DEFINITION INTEGER-ABS) (:DEFINITION O<)) Subgoal 2 Subgoal 1 Q.E.D. That completes the proof of the measure theorem for FIB. Thus, we admit this function under the principle of definition. We observe that the type of FIB is described by the theorem (AND (INTEGERP (FIB N)) (<= 0 (FIB N))). We used primitive type reasoning. Computing the guard conjecture for FIB.... The non-trivial part of the guard conjecture for FIB, given the :compound- recognizer rules NATP-COMPOUND-RECOGNIZER and ZP-COMPOUND-RECOGNIZER, the :executable-counterpart of EQLABLEP, primitive type reasoning and the :type-prescription rule FIB, is Goal (IMPLIES (AND (NATP N) (NOT (ZP N)) (NOT (EQL N 1))) (NATP (+ -2 N))). Goal' Goal'' Q.E.D. That completes the proof of the guard theorem for FIB. FIB is compliant with Common Lisp. Summary Form: ( DEFUN FIB ...) Rules: ((:COMPOUND-RECOGNIZER NATP-COMPOUND-RECOGNIZER) (:COMPOUND-RECOGNIZER ZP-COMPOUND-RECOGNIZER) (:DEFINITION ACL2-COUNT) (:DEFINITION EQL) (:DEFINITION INTEGER-ABS) (:DEFINITION NATP) (:DEFINITION NOT) (:DEFINITION O-FINP) (:DEFINITION O<) (:EXECUTABLE-COUNTERPART EQLABLEP) (:EXECUTABLE-COUNTERPART TAU-SYSTEM) (:EXECUTABLE-COUNTERPART UNARY--) (:FAKE-RUNE-FOR-TYPE-SET NIL) (:REWRITE DISTRIBUTIVITY-OF-MINUS-OVER-+) (:TYPE-PRESCRIPTION FIB)) Splitter rules (see :DOC splitter): if-intro: ((:DEFINITION ACL2-COUNT) (:DEFINITION INTEGER-ABS) (:DEFINITION O<)) FIB ACL2 !>>(WITH-OUTPUT :OFF SUMMARY (PROGN (COMP 'FIB) (VALUE-TRIPLE T))) ACL2 !>>>(COMP 'FIB) ACL2 !>>>(VALUE-TRIPLE T) T ACL2 !>>(DEFUN FIB2 (N) (DECLARE (XARGS :GUARD (NATP N))) (IF (= N 40) 102334155 (FIB N))) Since FIB2 is non-recursive, its admission is trivial. We observe that the type of FIB2 is described by the theorem (AND (INTEGERP (FIB2 N)) (<= 0 (FIB2 N))). We used the :type-prescription rule FIB. Computing the guard conjecture for FIB2.... The guard conjecture for FIB2 is trivial to prove, given the :compound- recognizer rule NATP-COMPOUND-RECOGNIZER. FIB2 is compliant with Common Lisp. Summary Form: ( DEFUN FIB2 ...) Rules: ((:COMPOUND-RECOGNIZER NATP-COMPOUND-RECOGNIZER) (:TYPE-PRESCRIPTION FIB)) FIB2 ACL2 !>>(MEMOIZE 'FIB :INVOKE 'FIB2) ACL2 Error in ( TABLE MEMOIZE-TABLE ...): Illegal attempt to set memoize- table: The following event must be admitted (possibly with differing name or macro) before memoizing function FIB with :INVOKE value FIB2. See :DOC memoize. (DEFTHM |FIB-is-FIB2| (EQUAL (FIB N) (FIB2 N)) :RULE-CLASSES NIL) ACL2 Error [Failure] in ( TABLE MEMOIZE-TABLE ...): See :DOC failure. ACL2 !>>(DEFTHM |FIB-is-FIB2| (EQUAL (FIB N) (FIB2 N)) :RULE-CLASSES NIL) Q.E.D. Summary Form: ( DEFTHM |FIB-is-FIB2| ...) Rules: ((:DEFINITION FIB2) (:EXECUTABLE-COUNTERPART EQUAL) (:EXECUTABLE-COUNTERPART FIB)) |FIB-is-FIB2| ACL2 !>>(MEMOIZE 'FIB :INVOKE 'FIB2) FIB ACL2 !>>(FIB 40) 102334155 ACL2 !>>(FIB 41) 165580141 ACL2 !>>(DEFUN F2 (X) (DECLARE (XARGS :GUARD (CONSP X))) (CAR X)) Since F2 is non-recursive, its admission is trivial. We could deduce no constraints on the type of F2. Computing the guard conjecture for F2.... The guard conjecture for F2 is trivial to prove. F2 is compliant with Common Lisp. Summary Form: ( DEFUN F2 ...) Rules: NIL F2 ACL2 !>>(F2 3) ACL2 Error [Evaluation] in TOP-LEVEL: The guard for the function call (F2 X), which is (CONSP X), is violated by the arguments in the call (F2 3). See :DOC set-guard-checking for information about suppressing this check with (set-guard-checking :none), as recommended for new users. To debug see :DOC print-gv, see :DOC trace, and see :DOC wet. ACL2 !>>(DEFUN G2 (X) (DECLARE (XARGS :GUARD (LISTP X))) (CAR X)) Since G2 is non-recursive, its admission is trivial. We could deduce no constraints on the type of G2. Computing the guard conjecture for G2.... The guard conjecture for G2 is trivial to prove. G2 is compliant with Common Lisp. Summary Form: ( DEFUN G2 ...) Rules: NIL G2 ACL2 !>>(DEFTHM |F-is-G| (EQUAL (F2 X) (G2 X)) :RULE-CLASSES NIL) Q.E.D. Summary Form: ( DEFTHM |F-is-G| ...) Rules: ((:DEFINITION F2) (:DEFINITION G2)) |F-is-G| ACL2 !>>(MEMOIZE 'F2 :INVOKE 'G2) ACL2 Error in ( TABLE MEMOIZE-TABLE ...): Illegal attempt to set memoize- table: The following event must be admitted (possibly with differing name or macro) before memoizing function F2 with :INVOKE value G2. See :DOC memoize. (VERIFY-GUARD-IMPLICATION F2 G2) ACL2 Error [Failure] in ( TABLE MEMOIZE-TABLE ...): See :DOC failure. ACL2 !>>(VERIFY-GUARD-IMPLICATION F2 G2) Q.E.D. Summary Form: ( DEFTHM |F2-guard-implies-G2-guard| ...) Rules: NIL Summary Form: ( MAKE-EVENT (LET ...)) Rules: NIL |F2-guard-implies-G2-guard| ACL2 !>>(MEMOIZE 'F2 :INVOKE 'G2) F2 ACL2 !>>(TRACE$ G2) ((G2)) ACL2 !>>(F2 3) ACL2 Error [Evaluation] in TOP-LEVEL: The guard for the function call (F2 X), which is (CONSP X), is violated by the arguments in the call (F2 3). See :DOC set-guard-checking for information about suppressing this check with (set-guard-checking :none), as recommended for new users. To debug see :DOC print-gv, see :DOC trace, and see :DOC wet. ACL2 !>>(F2 '(A B C)) 1> (G2 (A B C)) <1 (G2 A) A ACL2 !>>(DEFN F3 (X Y) (DECLARE (TYPE RATIONAL X Y)) (+ X Y)) Since F3 is non-recursive, its admission is trivial. We observe that the type of F3 is described by the theorem (ACL2-NUMBERP (F3 X Y)). We used primitive type reasoning. Computing the guard conjecture for F3.... The guard conjecture for F3 is trivial to prove. F3 is compliant with Common Lisp. Summary Form: ( DEFUN F3 ...) Rules: ((:FAKE-RUNE-FOR-TYPE-SET NIL)) F3 ACL2 !>>(DEFN G3 (X Y) (DECLARE (TYPE RATIONAL X Y)) (+ 1 -1 X Y)) Since G3 is non-recursive, its admission is trivial. We observe that the type of G3 is described by the theorem (ACL2-NUMBERP (G3 X Y)). We used primitive type reasoning. Computing the guard conjecture for G3.... The guard conjecture for G3 is trivial to prove, given primitive type reasoning. G3 is compliant with Common Lisp. Summary Form: ( DEFUN G3 ...) Rules: ((:FAKE-RUNE-FOR-TYPE-SET NIL)) G3 ACL2 !>>(DEFTHM |F3-is-G3| (EQUAL (F3 X Y) (G3 X Y)) :RULE-CLASSES NIL) Q.E.D. Summary Form: ( DEFTHM |F3-is-G3| ...) Rules: ((:DEFINITION F3) (:DEFINITION G3) (:EXECUTABLE-COUNTERPART TAU-SYSTEM)) |F3-is-G3| ACL2 !>>(WITH-OUTPUT :OFF :ALL (MEMOIZE 'F3 :INVOKE 'G3 :COMMUTATIVE T)) T ACL2 !>>(TRACE$ G3) ((G3)) ACL2 !>>(F3 3 4) 1> (G3 3 4) <1 (G3 7) 7 ACL2 !>>Bye.