Logic 如何自动证明两个一阶公式是等价的？

自动证明两个一阶公式F和G等价的最佳方法是什么

与“完整”一阶公式相比，这些公式有一些限制：

无量词

无功能

隐式普遍量化

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我可以用子句范式转换这些公式，我有统一文本的例程。然而，我不确定如何继续，以及这个问题是否是可判定的。

如前所述，要证明两个公式都是闭合的（普遍量化的）公式，需要证明F=>G和G=>F。要证明这两个公式中的每一个，可以使用各种计算。我将描述[解析演算]：

• 否定这个猜想，使F=>G变成F&-G
• 转换成CNF
• 运行解析过程
• 如果你得到一个空子句，你已经证明了原来的猜想F=>G。如果搜索饱和，不能再得到新子句，这个猜想就不成立了

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，来自G的所有原子公式将是仅应用于skolem常量的谓词符号。所以解析过程只会产生替换，要么将变量映射到其他变量，要么将变量映射到那些skolem常量。这意味着它只能导出有限数量的不同文字，因此解析过程将始终停止-它将是可判定的

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您还可以为此目的使用自动化工具，该工具将为您完成所有工作。我用它来解决这些问题。作为输入语言，我使用的语言是，人类很容易读/写

举个例子：输入文件：

fof(my_formula_name, conjecture, (![X]: p(X)) <=> (![Y]: p(Y)) ).

（

-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

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这个问题取决于你对公式的限制程度（我不记得细节）。为了证明F和g是等价的，你需要证明F意味着g和g意味着F。为了证明你试图用荒谬来证明它，通过试图证明F而不是（g）并且失败。。。然后探索这个公式的后果，同时试图证明它是错误的。该方法被称为类似“反驳树”或“真理树”的东西，因为它是无函数的，并且完全通用，将您置于一个可判定的片段中：

# 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

# Proof found!
# SZS status Theorem