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test.sh
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test.sh
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#!/bin/sh
set -e
if [ $# != 1 ]; then
echo "Usage: ${0##*/} PROGRAM" >&2
exit 1
fi
PROGRAM="$1"
t(){
printf "%s...\\n" "$1"
tee "$1" | time -p "$PROGRAM"
printf "Done\\n"
}
cat << EOF | t positive-implicational.pa
axiomatization
implies :: "prop => (prop => prop)" (infixr "-->" 25)
where
MP: [| "A --> B"; "A" |] ==> "B"
allow_deduction implies
lemma A1: "A --> (B --> A)"
proof
{
assume a: "A"
{
assume "B"
have "A" by a
}
}
show by this
qed
lemma A2: "(A --> (B --> C)) --> ((A --> B) --> (A --> C))"
proof
{
assume abc: "A --> (B --> C)"
{
assume ab: "A --> B"
{
assume "A"
have b: "B" by MP(ab, this)
have bc: "B --> C" by MP(abc, assm)
have "C" by MP(bc, b)
}
}
}
show by this
qed
lemma a_impl_a: "A --> A"
proof
{
assume "A"
}
show by this
qed
lemma syllogism: "(A --> B) --> ((B --> C) --> (A --> C))"
proof
{
assume ab: "A --> B"
{
assume bc: "B --> C"
{
assume "A"
have "B" by MP(ab, this)
have "C" by MP(bc, this)
}
}
}
show by this
qed
lemma swap_premises: "(A --> (B --> C)) --> (B --> (A --> C))"
proof
{
assume abc: "A --> (B --> C)"
{
assume b: "B"
{
assume "A"
have "B --> C" by MP(abc, this)
have "C" by MP(this, b)
}
}
}
show by this
qed
lemma reverse_syllogism: "(B --> C) --> ((A --> B) --> (A --> C))" by MP(swap_premises, syllogism)
EOF
cat positive-implicational.pa - << EOF | t positive.pa
axiomatization
conj :: "prop => (prop => prop)" (infixl "&" 35) and
disj :: "prop => (prop => prop)" (infixl "|" 30)
where
A3: "A & B --> A" and
A4: "A & B --> B" and
A5: "A --> (B --> A & B)" and
A6: "A --> A | B" and
A7: "B --> A | B" and
A8: "(A --> C) --> ((B --> C) --> (A | B --> C))"
EOF
cat positive.pa - << EOF | t minimal.pa
axiomatization
Not :: "prop => prop" (prefix "~" 40)
where
A9: "(A --> B) --> ((A --> ~B) --> ~A)"
lemma contraposition: "(A --> B) --> (~B --> ~A)"
proof
{
assume "A --> B"
have "(A --> ~B) --> ~A" by MP(A9, this)
have "~B --> ~A" by MP(MP(syllogism, A1), this)
}
show by this
qed
lemma intro_2_neg: "A --> ~~A"
proof
{
assume "A"
have "~A --> A" by MP(A1, this)
have "~~A" by MP(MP(A9, this), a_impl_a)
}
show by this
qed
lemma elim_3_neg: "~~~A --> ~A"
proof
have a: "(A --> ~~~A) --> ~A" by MP(A9, intro_2_neg)
{
assume "~~~A"
have "A --> ~~~A" by MP(A1, this)
have "~A" by MP(a, this)
}
show by this
qed
lemma ex_falso_not: "A --> (~A --> ~B)"
proof
{
assume "A"
have ba: "B --> A" by MP(A1, this)
{
assume "~A"
have "B --> ~A" by MP(A1, this)
have "~B" by MP(MP(A9, ba), this)
}
}
show by this
qed
EOF
cat positive.pa - << EOF | t alt-minimal.pa
axiomatization
Not :: "prop => prop" (prefix "~" 40)
where
contraposition: "(A --> B) --> (~B --> ~A)" and
intro_2_neg: "A --> ~~A"
lemma a_impl_not_b_impl_b_impl_not_a: "(A --> ~B) --> (B --> ~A)"
proof
have "B --> ~~B" by intro_2_neg
have a: "(~~B --> ~A) --> (B --> ~A)" by MP(syllogism, this)
have "(A --> ~B) --> (~~B --> ~A)" by contraposition
show by MP(MP(syllogism, this), a)
qed
lemma not_a_impl_b_impl_not_b_impl_a: "A --> (~B --> ~(A --> B))"
proof
have a: "A --> ((A --> B) --> B)" by MP(swap_premises, a_impl_a)
have "((A --> B) --> B) --> (~B --> ~(A --> B))" by contraposition
show by MP(MP(syllogism, a), this)
qed
lemma a_impl_not_a_impl_not_a: "(A --> ~A) --> ~A"
proof
have a: "A --> (~~A --> ~(A --> ~A))" by not_a_impl_b_impl_not_b_impl_a
have "A --> ~~A" by intro_2_neg
have "A --> ~(A --> ~A)" by MP(MP(A2, a), this)
show by MP(a_impl_not_b_impl_b_impl_not_a, this)
qed
lemma A9: "(A --> B) --> ((A --> ~B) --> ~A)"
proof
{
assume "A --> B"
have "(B --> ~A) --> (A --> ~A)" by MP(syllogism, this)
have "(A --> ~B) --> (A --> ~A)" by MP(MP(syllogism, a_impl_not_b_impl_b_impl_not_a), this)
have "(A --> ~B) --> ~A" by MP(MP(syllogism, this), a_impl_not_a_impl_not_a)
}
show by this
qed
EOF
# Эквивалентность minimal и alt-minimal доказана
# Если к positive добавить константу False и никаких аксиом, получится логика, эквивалентная minimal, я не буду это доказывать
cat minimal.pa - << EOF | t intuitionistic.pa
axiomatization where
ex_falso: "A --> (~A --> B)"
lemma plan_b: "A | B --> (~A --> B)" by MP(MP(A8, ex_falso), A1)
lemma not_a_or_b_impl_a_impl_b: "~A | B --> (A --> B)" by MP(MP(A8, MP(swap_premises, ex_falso)), A1)
EOF
# Будем называть vml2010 конспекты лекций 1 - 8 курса "Введение в математическую логику" Л. Д. Беклемишева ( http://lpcs.math.msu.su/vml2010 ). Также будем называть vml2010 формулировку классического исчисления высказываний, введённую в этих конспектах (10 аксиом A1 - A10 и MP)
cat intuitionistic.pa - << EOF | t alt-classical.pa
# Альтернативная к vml2010 формулировка классического исчисления
axiomatization where
excluded_middle: "A | ~A"
lemma A10: "~~A --> A"
proof
have "A | ~A --> (~~A --> A)" by MP(MP(A8, A1), ex_falso)
show by MP(this, excluded_middle)
qed
EOF
cat minimal.pa - << EOF | t classical.pa
# vml2010
axiomatization where
A10: "~~A --> A"
# Это правило формализует "классическое рассуждение", противопоставленное интуиционистскому. Это формализация рассуждения "Если sqrt(2)^sqrt(2) рационально, то существуют рациональные числа вида a^b, где a и b иррациональны. Если иррационально - то тоже существуют. Значит, они существуют", которое обычно приводят в пример "неправильного" рассуждения интуционисты
lemma classical_reasoning: "(A --> B) --> ((~A --> B) --> B)"
proof
{
assume "A --> B"
have "~B --> ~A" by MP(contraposition, this)
have a: "(~B --> ~~A) --> ~~B" by MP(A9, this)
{
assume "~A --> B"
have "~B --> ~~A" by MP(contraposition, this)
have "~~B" by MP(a, this)
have "B" by MP(A10, this)
}
}
show by this
qed
lemma ex_falso: "A --> (~A --> B)"
proof
have a: "A --> (~B --> A)" by A1
have "(~B --> A) --> ((~B --> ~A) --> ~~B)" by A9
have "A --> ((~B --> ~A) --> ~~B)" by MP(MP(syllogism, a), this)
have b: "(~B --> ~A) --> (A --> ~~B)" by MP(swap_premises, this)
have "~A --> (~B --> ~A)" by A1
have c: "~A --> (A --> ~~B)" by MP(MP(syllogism, this), b)
have "~~B --> B" by A10
have "(A --> ~~B) --> (A --> B)" by MP(reverse_syllogism, this)
have "~A --> (A --> B)" by MP(MP(syllogism, c), this)
show by MP(swap_premises, this)
qed
lemma excluded_middle: "A | ~A" by MP(MP(classical_reasoning, A6), A7)
lemma plan_b: "A | B --> (~A --> B)" by MP(MP(A8, ex_falso), A1)
EOF
# Эквивалентность classical и alt-classical доказана
cat classical.pa - << EOF | t false.pa
axiomatization
False :: "prop" ("FALSE")
where
not_false: "~ False"
lemma false_impl_a: "False --> A"
proof
{
assume "False"
have "A" by MP(MP(ex_falso, this), not_false)
}
show by this
qed
lemma contradiction2: "(A --> False) --> ~A"
proof
{
assume "A --> False"
have "A --> ~False" by MP(A1, not_false)
have "~A" by MP(MP(A9, assm), this)
}
show by this
qed
lemma contradiction: "(~A --> False) --> A"
proof
{
assume "~A --> False"
have "~~A" by MP(contradiction2, this)
have "A" by MP(A10, this)
}
show by this
qed
EOF
cat false.pa - << EOF | t modal-k.pa
# Нельзя использовать дедукцию из-за Necessitation Rule
deny_deduction
axiomatization
box :: "prop => prop" (prefix "[]" 50)
where
# Distribution Axiom
dist: "[](A --> B) --> ([]A --> []B)" and
# Necessitation Rule
nec: "A" ==> "[]A"
EOF
cat false.pa - << EOF | t fol.pa
# Типа FOL, с функциями высшего порядка
axiomatization
All :: "('a => prop) => prop" (binder "!") and
Ex :: "('a => prop) => prop" (binder "?")
where
spec: "All A --> A t" and
exI: "A t --> Ex A"
allow_fixing All
allow_obtaining Ex
EOF
cat fol.pa - << EOF | t eq.pa
axiomatization
eq :: "'a => ('a => prop)" (infix "=" 50)
where
refl: "a = a" and
subst: "s = t --> (P s --> P t)" and
ext: "(! x. f x = g x) --> f = g"
lemma eq_commute: "(a::'a) = b --> b = a"
proof
{
assume "a = b"
have "a = b --> ((%x. x = a) a --> (%x. x = a) b)" by subst
have "a = b --> (a = a --> b = a)" by this
have "b = a" by MP(MP(this, assm), refl)
}
show by this
qed
lemma trans: "(r::'a) = s --> (s = t --> r = t)"
proof
have "s = r --> ((%x. x = t) s --> (%x. x = t) r)" by subst
have "s = r --> (s = t --> r = t)" by this
show by MP(MP(syllogism, eq_commute), this)
qed
lemma arg_cong: "(x::'a) = y --> ((f x)::'b) = f y"
proof
{
assume "x = y"
have a: "y = x" by MP(eq_commute, this)
have "y = x --> ((%z. f z = f y) y --> (%z. f z = f y) x)" by subst
have "y = x --> (f y = f y --> f x = f y)" by this
have "f x = f y" by MP(MP(this, a), refl)
}
show by this
qed
lemma fun_cong: "(f :: 'a => 'b) = g --> f x = g x"
proof
have "f = g --> (%h. h x) f = (%h. h x) g" by arg_cong
show by this
qed
axiomatization
comp :: "('b => 'c) => (('a => 'b) => ('a => 'c))"
where
comp_def: "comp f g x = f (g x)"
lemma comp_assoc: "comp (comp (f :: 'c => 'd) (g :: 'b => 'c)) (h :: 'a => 'b) = comp f (comp g h)"
proof
{
fix x :: "'a"
have "comp (comp f g) h x = comp f g (h x)" by comp_def
have "comp (comp f g) h x = f (g (h x))" by MP(MP(trans, this), comp_def)
have "comp (comp f g) h x = f (comp g h x)" by MP(MP(trans, this), MP(arg_cong, MP(eq_commute, comp_def)))
have "comp (comp f g) h x = comp f (comp g h) x" by MP(MP(trans, this), MP(eq_commute, comp_def))
}
show by MP(ext, this)
qed
EOF
cat eq.pa - << EOF | t pa.pa
typedecl i
axiomatization
zero :: "i" ("0") and
succ :: "i => i" (prefix "SUCC " 999)
where
PA1: "~ succ a = 0" and
PA2: "a = 0 | ? b. a = succ b" and
PA3: "succ a = succ b --> a = b" and
PA4: "P 0 --> ((! x. P x --> P (succ x)) --> P x)"
axiomatization
plus :: "i => (i => i)" (infixl "+" 65)
where
plus_zero: "a + 0 = a" and
plus_succ: "a + succ b = succ (a + b)"
lemma zero_plus_a: "0 + a = a"
proof
have base: "(%x. 0 + x = x) 0" by plus_zero
{
fix x :: "i"
have a: "0 + succ x = succ (0 + x)" by plus_succ
{
assume "0 + x = x"
have "succ (0 + x) = succ x" by MP(arg_cong, assm)
have "0 + succ x = succ x" by MP(MP(trans, a), this)
}
}
show by MP(MP(PA4, base), this)
qed
lemma succ_a_plus_b: "succ a + b = succ (a + b)"
proof
have "succ a + 0 = succ a" by plus_zero
have base: "(%x. succ a + x = succ (a + x)) 0" by MP(MP(trans, this), MP(arg_cong, MP(eq_commute, plus_zero)))
{
fix x :: "i"
{
assume "succ a + x = succ (a + x)"
have "succ a + succ x = succ (succ a + x)" by plus_succ
have "succ a + succ x = succ (succ (a + x))" by MP(MP(trans, this), MP(arg_cong, assm))
have "succ a + succ x = succ (a + succ x)" by MP(MP(trans, this), MP(arg_cong, MP(eq_commute, plus_succ)))
}
}
show by MP(MP(PA4, base), this)
qed
lemma plus_commute: "a + b = b + a"
proof
have base: "(%x. x + b = b + x) 0" by MP(MP(trans, zero_plus_a), MP(eq_commute, plus_zero))
{
fix x :: "i"
{
assume "x + b = b + x"
have "succ x + b = succ (x + b)" by succ_a_plus_b
have "succ x + b = succ (b + x)" by MP(MP(trans, this), MP(arg_cong, assm))
have "succ x + b = b + succ x" by MP(MP(trans, this), MP(eq_commute, plus_succ))
}
}
show by MP(MP(PA4, base), this)
qed
axiomatization
le :: "i => (i => prop)" (infix "<=" 50)
where
LE1: "a <= 0 --> a = 0" and
LE2: "a = 0 --> a <= 0" and
LE3: "a <= succ b --> a = succ b | a <= b" and
LE4: "a = succ b --> a <= succ b" and
LE5: "a <= b --> a <= succ b"
axiomatization
less :: "i => (i => prop)" (infix "<" 50)
where
L1: "a < b --> a <= b" and
L2: "a < b --> ~(a = b)" and
L3: "a <= b --> (~(a = b) --> a < b)"
lemma less_zero: "~ a < 0"
proof
{
assume "a < 0"
have "a <= 0" by MP(L1, this)
have "a = 0" by MP(LE1, this)
have "False" by MP(MP(ex_falso, this), MP(L2, assm))
}
show by MP(contradiction2, this)
qed
lemma less_succ: "a < succ b --> a <= b"
proof
{
assume "a < succ b"
have "a <= succ b" by MP(L1, assm)
have a: "a = succ b | a <= b" by MP(LE3, this)
have "~(a = succ b)" by MP(L2, assm)
have "a <= b" by MP(MP(plan_b, a), this)
}
show by this
qed
lemma le: "a <= b --> a < b | a = b"
proof
{
assume a: "a <= b"
have switch: "b = 0 | ? x. b = succ x" by PA2
{
assume "b = 0"
have "a <= 0" by MP(MP(subst, this), a)
have "a = 0" by MP(LE1, this)
have "a = b" by MP(MP(trans, this), MP(eq_commute, assm))
have "a < b | a = b" by MP(A7, this)
}
note casea = this
{
assume "? x. b = succ x"
{
obtain x :: "i" where "b = succ x" by this
have "a <= succ x" by MP(MP(subst, this), a)
have b: "a = succ x | a <= x" by MP(LE3, this)
{
assume "a = succ x"
have "a = b" by MP(MP(trans, this), MP(eq_commute, obtaining))
have "a < b | a = b" by MP(A7, this)
}
note casea2 = this
{
assume "~(a = succ x)"
have "a <= x" by MP(MP(plan_b, b), this)
have "a <= succ x" by MP(LE5, this)
have "a < succ x" by MP(MP(L3, this), assm)
have "a < b" by MP(MP(subst, MP(eq_commute, obtaining)), this)
have "a < b | a = b" by MP(A6, this)
}
note caseb2 = this
have "a < b | a = b" by MP(MP(classical_reasoning, casea2), caseb2)
}
}
note caseb = this
have "a < b | a = b" by MP(MP(MP(A8, casea), caseb), switch)
}
show by this
qed
lemma full_induction: "(! x. (! y. y < x --> P y) --> P x) --> P a"
proof
{
assume "! x. (! y. y < x --> P y) --> P x"
{
fix t :: "i"
have "t < 0 --> P t" by MP(MP(swap_premises, ex_falso), less_zero)
}
have base: "(%z. (! t. t < z --> P t)) 0" by this
{
fix z :: "i"
have a: "(! y. y < z --> P y) --> P z" by MP(spec, assm)
{
assume "! t. t < z --> P t"
have pz: "P z" by MP(a, this)
{
fix t :: "i"
have casea: "t < z --> P t" by MP(spec, assm)
{
assume "t < succ z"
have "t <= z" by MP(less_succ, this)
have switch: "t < z | t = z" by MP(le, this)
{
assume "t = z"
have "P t" by MP(MP(subst, MP(eq_commute, this)), pz)
}
note caseb = this
have "P t" by MP(MP(MP(A8, casea), caseb), switch)
}
have "t < succ z --> P t" by this
}
have "! t. t < succ z --> P t" by this
}
have "(! t. t < z --> P t) --> (! t. t < succ z --> P t)" by this
}
have "! t. t < a --> P t" by MP(MP(PA4, base), this)
have "P a" by MP(MP(spec, assm), this)
}
show by this
qed
axiomatization
one :: "i" ("1") and
two :: "i" ("2") and
three :: "i" ("3") and
four :: "i" ("4")
where
one_def: "1 = succ 0" and
two_def: "2 = succ 1" and
three_def: "3 = succ 2" and
four_def: "4 = succ 3"
lemma two_plus_two: "2 + 2 = 4"
proof
have "2 + 2 = 2 + succ 1" by MP(arg_cong, two_def)
have "2 + 2 = succ (2 + 1)" by MP(MP(trans, this), plus_succ)
have "2 + 2 = succ (2 + succ 0)" by MP(MP(trans, this), MP(arg_cong, MP(arg_cong, one_def)))
have "2 + 2 = succ (succ (2 + 0))" by MP(MP(trans, this), MP(arg_cong, plus_succ))
have "2 + 2 = succ (succ 2)" by MP(MP(trans, this), MP(arg_cong, MP(arg_cong, plus_zero)))
have "2 + 2 = succ 3" by MP(MP(trans, this), MP(arg_cong, MP(eq_commute, three_def)))
show by MP(MP(trans, this), MP(eq_commute, four_def))
qed
axiomatization
times :: "i => (i => i)" (infixl "*" 70)
where
times_zero: "a * 0 = 0" and
times_succ: "a * succ b = a * b + a"
EOF
cat eq.pa - << EOF | t group.pa
typedecl i
axiomatization
e :: "i" and
times :: "i => (i => i)" (infixl "*" 70) and
inv :: "i => i"
where
assoc: "(a * b) * c = a * (b * c)" and
e1: "a * e = a" and
e2: "e * a = a" and
inv1: "a * inv a = e" and
inv2: "inv a * a = e"
lemma one_e: "i * a = e --> a * i = e --> j * a = e --> a * j = e --> i = j"
proof
{
assume iae: "i * a = e"
{
assume aie: "a * i = e"
{
assume jae: "j * a = e"
{
assume aje: "a * j = e"
have "i = i * e" by MP(eq_commute, e1)
have "i = i * (a * j)" by MP(MP(trans, this), MP(arg_cong, MP(eq_commute, aje)))
have "i = (i * a) * j" by MP(MP(trans, this), MP(eq_commute, assoc))
have "i = e * j" by MP(MP(trans, this), MP(fun_cong, MP(arg_cong, iae)))
have "i = j" by MP(MP(trans, this), e2)
}
}
}
}
show by this
qed
EOF