自动证明两个一阶公式F和G是等价的最佳方法是什么?
与“完整”一阶公式相比,公式有一些限制:
我可以用正则形式的子句转换那些公式,并且我有文本统一的例程。但是,我不确定如何继续,如果这个问题是可判定的。
答案 0 :(得分:3)
如上所述,为了证明 F< => G 两者都是封闭的(普遍量化的)公式,你需要证明 F => G 以及 G => ˚F。为了证明这两个公式中的每一个,你可以使用各种计算。我将描述[解析微积分]:
在您的条件下,来自 F 的所有原子公式将是仅应用于变量的谓词符号,而来自 G 的所有原子公式将是仅应用于skolem常量的谓词符号。因此,解析过程只会产生替换,将变量映射到其他变量,或变量映射到那些skolem常量。这意味着它只能导出有限数量的不同文字,因此解析程序将始终停止 - 它将是可判定的。
您也可以使用自动化工具来完成所有适合您的工作。我使用The E Theorem Prover来解决这些问题。作为输入语言,我使用The TPTP Problem Library语言,这对人类来说很容易读/写。
举个例子:输入文件:
fof(my_formula_name, conjecture, (![X]: p(X)) <=> (![Y]: p(Y)) ).
然后我跑
eprover --tstp-format -xAuto -tAuto myfile
(-tAuto
和-xAuto
执行一些自动配置,在您的情况下很可能不需要),结果是
# Garbage collection reclaimed 59 unused term cells.
# Auto-Ordering is analysing problem.
# Problem is type GHNFGFFSF00SS
# Auto-mode selected ordering type KBO6
# Auto-mode selected ordering precedence scheme <invfreq>
# Auto-mode selected weight ordering scheme <precrank20>
#
# Auto-Heuristic is analysing problem.
# Problem is type GHNFGFFSF00SS
# Auto-Mode selected heuristic G_E___107_C41_F1_PI_AE_Q4_CS_SP_PS_S0Y
# and selection function SelectMaxLComplexAvoidPosPred.
#
# No equality, disabling AC handling.
#
# Initializing proof state
#
#cnf(i_0_2,negated_conjecture,(~p(esk1_0)|~p(esk2_0))).
#
#cnf(i_0_1,negated_conjecture,(p(X1)|p(X2))).
# Presaturation interreduction done
#
#cnf(i_0_2,negated_conjecture,(~p(esk1_0)|~p(esk2_0))).
#
#cnf(i_0_1,negated_conjecture,(p(X2)|p(X1))).
#
#cnf(i_0_3,negated_conjecture,(p(X3))).
# Proof found!
# SZS status Theorem
# Parsed axioms : 1
# Removed by relevancy pruning : 0
# Initial clauses : 2
# Removed in clause preprocessing : 0
# Initial clauses in saturation : 2
# Processed clauses : 5
# ...of these trivial : 0
# ...subsumed : 0
# ...remaining for further processing : 5
# Other redundant clauses eliminated : 0
# Clauses deleted for lack of memory : 0
# Backward-subsumed : 1
# Backward-rewritten : 1
# Generated clauses : 4
# ...of the previous two non-trivial : 4
# Contextual simplify-reflections : 0
# Paramodulations : 2
# Factorizations : 2
# Equation resolutions : 0
# Current number of processed clauses : 1
# Positive orientable unit clauses : 1
# Positive unorientable unit clauses: 0
# Negative unit clauses : 0
# Non-unit-clauses : 0
# Current number of unprocessed clauses: 0
# ...number of literals in the above : 0
# Clause-clause subsumption calls (NU) : 0
# Rec. Clause-clause subsumption calls : 0
# Unit Clause-clause subsumption calls : 1
# Rewrite failures with RHS unbound : 0
# Indexed BW rewrite attempts : 4
# Indexed BW rewrite successes : 4
# Unification attempts : 12
# Unification successes : 9
# Backwards rewriting index : 2 leaves, 1.00+/-0.000 terms/leaf
# Paramod-from index : 1 leaves, 1.00+/-0.000 terms/leaf
# Paramod-into index : 1 leaves, 1.00+/-0.000 terms/leaf
最重要的行是
# Proof found!
# SZS status Theorem