Proof of proveit.logic.booleans.conjunction.and_if_both theorem

import proveit
from proveit import A, B
from proveit.logic.booleans.conjunction import true_and_true
theory = proveit.Theory() # the theorem's theory

%proving and_if_both

With these allowed/disallowed theorem/theory presumptions (e.g., to avoid circular dependencies), we begin our proof of
and_if_both:
(see dependencies)

and_if_both may now be readily provable (assuming required theorems are usable).  Simply execute "%qed".

AeqT = A.evaluation(assumptions=[A])

AeqT:  ⊢  

BeqT = B.evaluation(assumptions=[B])

BeqT:  ⊢  

true_and_true

 ⊢  

AandT = AeqT.sub_left_side_into(true_and_true.inner_expr().operands[0],
                                auto_simplify=False)

AandT:  ⊢  

AandB = BeqT.sub_left_side_into(AandT, assumptions=[A, B], 
                                auto_simplify=False)

AandB: ,  ⊢  

# To do this manually, execute AandB.generalize((A, B), conditions=[A, B]).
# But this can be figured out via automation.
%qed

proveit.logic.booleans.conjunction.and_if_both has been proven.

	step type	requirements	statement
0	generalization	1	⊢
1	instantiation	4, 2, 3	,  ⊢
	: , :
2	instantiation	4, 5, 6	⊢
	: , :
3	assumption		⊢
4	theorem		⊢
	proveit.logic.equality.substitute_truth
5	theorem		⊢
	proveit.logic.booleans.conjunction.true_and_true
6	assumption		⊢