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Zhangir Azerbayev
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horizon := 1;;
thm `;
!R. (!x. R x x) /\ (!x y z. R x y /\ R y z ==> R x z)
==> ((!m n. m <= n ==> R m n) <=> (!n. R n (SUC n)))
proof
let R be num->num->bool;
assume !x. R x x [1];
assume !x y z. R x y /\ R y z ==> R x z [2];
!n. n <= SUC n by ARITH_TAC;
(!m n. m <= n ==> R m n) ==> (!n. R n (SUC n)) [3] by SIMP_TAC;
now
assume !n. R n (SUC n) [4];
!m n d. n = m + d ==> R m (m + d)
proof
let m be num;
R m m by MESON_TAC,1;
R m (m + 0) [5] by REWRITE_TAC,ADD_CLAUSES;
!d. R m (m + d) ==> R m (m + SUC d)
proof
let d be num;
assume R m (m + d);
R m (SUC (m + d)) by MESON_TAC,2,4;
qed by REWRITE_TAC,ADD_CLAUSES;
!d. R m (m + d) by INDUCT_TAC,5;
!d n. n = m + d ==> R m (m + d)
by REWRITE_TAC,LEFT_FORALL_IMP_THM,EXISTS_REFL,ADD_CLAUSES;
qed by ONCE_REWRITE_TAC,SWAP_FORALL_THM;
thus !m n. m <= n ==> R m n by SIMP_TAC,LE_EXISTS,LEFT_IMP_EXISTS_THM;
end;
qed by EQ_TAC,3`;;
thm `;
!s. INFINITE s ==> ?x:A. x IN s
proof
let s be A->bool;
assume INFINITE s;
~(s = {}) by INFINITE_NONEMPTY;
consider x such that
~(x IN s <=> x IN {}) [1] by EXTENSION;
take x;
~(x IN {}) by NOT_IN_EMPTY;
qed by 1`;;
let NOT_EVEN = thm `;
!n. ~EVEN n <=> ODD n
proof
~EVEN 0 <=> ODD 0 [1] by EVEN,ODD;
!n. (~EVEN n <=> ODD n) ==> (~EVEN (SUC n) <=> ODD (SUC n))
by EVEN,ODD;
qed by 1,INDUCT_TAC`;;