Theorem: Watson's Hypergeometric Theorem

:$\ds \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1} = \dfrac {\map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 + c} \map \Gamma {\dfrac 1 2 \paren {1 + a + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma {\dfrac 1 2 \paren {1 + a} } \map \Gamma {\dfrac 1 2 \paren {1 + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } }$

Proof:
From Thomae's Transformation, we have:
{{begin-eqn}}
{{eqn | l = \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {e, f} } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma e \map \Gamma f \map \Gamma s } {\map \Gamma a \map \Gamma {s + b} \map \Gamma {s + c} } \map { {}_3 \operatorname F_2} { { {e - a, f - a, s} \atop {s + b, s + c} } \, \middle \vert \, 1}
      | c = Thomae's Transformation: $s = e + f - a - b - c$
}}
{{eqn | ll = \leadsto
      | l = \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\paren {\dfrac 1 2 \paren {1 - a - b} + c } } } {\map \Gamma a \map \Gamma {\paren {\dfrac 1 2 \paren {1 - a - b} + c } + b} \map \Gamma {\paren {\dfrac 1 2 \paren {1 - a - b} + c } + c} } \map { {}_3 \operatorname F_2} { { {\dfrac 1 2 \paren {a + b + 1} - a, 2 c - a, \paren {\dfrac 1 2 \paren {1 - a - b} + c } } \atop {\paren {\dfrac 1 2 \paren {1 - a - b} + c } + b, \paren {\dfrac 1 2 \paren {1 - a - b} + c } + c} } \, \middle \vert \, 1}
      | c = $s = \paren {\dfrac 1 2 \paren {1 - a - b} + c }$
}}
{{eqn | n= 1
      | r = \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma a \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } } \map { {}_3 \operatorname F_2} { { {\dfrac 1 2 \paren {-a + b + 1}, 2 c - a,\dfrac 1 2 \paren {1 - a - b} + c } \atop {\dfrac 1 2 \paren {1 - a + b} + c, \dfrac 1 2 \paren {1 - a - b} + 2 c } } \, \middle \vert \, 1}
      | c = simplifying
}}
{{end-eqn}}
The generalized hypergeometric function on the {{RHS}} can be summed using Dixon's Hypergeometric Theorem.
From Dixon's Hypergeometric Theorem, we have:
:$\ds \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1} = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} } $
Let:
{{begin-eqn}}
{{eqn | l = n
      | r = 2 c - a
      | c = 
}}
{{eqn | l = -x
      | r = \dfrac 1 2 \paren {-a + b + 1}
      | c = 
}}
{{eqn | l = -y
      | r = \dfrac 1 2 \paren {1 - a - b} + c
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = x + n+ 1
      | r = \dfrac 1 2 \paren {1 - a - b} + 2 c
      | c = 
}}
{{eqn | l = y + n + 1
      | r = \dfrac 1 2 \paren {1 - a + b} + c
      | c = 
}}
{{eqn | l = \dfrac n 2 + 1
      | r = c - \dfrac a 2 + 1
      | c = 
}}
{{eqn | l = x + y + \dfrac n 2 + 1
      | r = \dfrac a 2
      | c = 
}}
{{eqn | l = n + 1
      | r = 1 - a + 2 c
      | c = 
}}
{{eqn | l = x + y + n + 1
      | r = c
      | c = 
}}
{{eqn | l = x + \dfrac n 2 + 1
      | r = \dfrac 1 2 \paren {1 - b} + c
      | c = 
}}
{{eqn | l = y + \dfrac n 2 + 1
      | r = \dfrac 1 2 \paren {1 + b}
      | c = 
}}
{{end-eqn}}
Making these substitutions to the generalized hypergeometric function on the {{RHS}}, we have:
{{begin-eqn}}
{{eqn | l = \map { {}_3 \operatorname F_2} { { {n, -x, -y} \atop {x + n + 1, y + n + 1} } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma {x + n + 1} \map \Gamma {y + n + 1} \map \Gamma {\dfrac n 2 + 1} \map \Gamma {x + y + \dfrac n 2 + 1} } { \map \Gamma {n + 1} \map \Gamma {x + y + n + 1} \map \Gamma {x + \dfrac n 2 + 1} \map \Gamma {y + \dfrac n 2 + 1} }
      | c = Dixon's Hypergeometric Theorem: Before substitution
}}
{{eqn | n = 2
      | ll= \leadsto
      | l = \map { {}_3 \operatorname F_2} { { {2 c - a, \dfrac 1 2 \paren {-a + b + 1}, \paren {\dfrac 1 2 \paren {1 - a - b} + c } } \atop {\paren {\dfrac 1 2 \paren {1 - a - b} + 2 c }, \paren {\dfrac 1 2 \paren {1 - a - b} + c } } } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma {2 c - a + 1} \map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }
      | c = Dixon's Hypergeometric Theorem: After substitution
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn | l = \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma a \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } } \dfrac {\map \Gamma {\dfrac 1 2 \paren {1 - a - b} + 2 c } \map \Gamma {\dfrac 1 2 \paren {1 - a + b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma {2 c - a + 1} \map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }
      | c = from $(1)$ and $(2)$ above
}}
{{eqn | r = \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {2 c} \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma a } \dfrac {\map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma {2 c - a + 1} \map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }
      | c = $\map \Gamma {c + \dfrac 1 2 \paren {1 - a + b} }$ and $\map \Gamma {2 c + \dfrac 1 2 \paren {1 - a - b} }$ cancel
}}
{{eqn | n = 3
      | r = \dfrac {\map \Gamma {2 c} } {\map \Gamma a \map \Gamma {2 c - a + 1} } \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }
      | c = simplifying
}}
{{end-eqn}}
From Legendre's Duplication Formula, we have:
{{begin-eqn}}
{{eqn | l = \map \Gamma {2 c}
      | r = \dfrac {2^{2 c - 1} \map \Gamma c \map \Gamma {c + \dfrac 1 2} } {\map \Gamma {\dfrac 1 2} }
      | c = 
}}
{{eqn | l = \map \Gamma a
      | r = \dfrac {2^{a - 1} \map \Gamma {\dfrac a 2} \map \Gamma {\dfrac 1 2 \paren {1 + a} } } {\map \Gamma {\dfrac 1 2} }
      | c = 
}}
{{eqn | l = \map \Gamma {2 c - a + 1}
      | r = \dfrac {2^{2 c - a} \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {c - \dfrac a 2 + 1} } {\map \Gamma {\dfrac 1 2} }
      | c = 
}}
{{end-eqn}}
Therefore:
{{begin-eqn}}
{{eqn | l = \map { {}_3 \operatorname F_2} { { {a, b, c} \atop {\dfrac 1 2 \paren {a + b + 1}, 2 c } } \, \middle \vert \, 1}
      | r = \dfrac {\map \Gamma {2 c} } {\map \Gamma a \map \Gamma {2 c - a + 1} } \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }
      | c = From $(3)$ above: Before substitution
}}
{{eqn | r = \dfrac {\paren {\dfrac {2^{2 c - 1} \map \Gamma c \map \Gamma {c + \dfrac 1 2} } {\map \Gamma {\dfrac 1 2} } } } {\paren {\dfrac {2^{a - 1} \map \Gamma {\dfrac a 2} \map \Gamma {\dfrac 1 2 \paren {1 + a} } } {\map \Gamma {\dfrac 1 2} } } \paren {\dfrac {2^{2 c - a} \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {c - \dfrac a 2 + 1} } {\map \Gamma {\dfrac 1 2} } } } \dfrac {\map \Gamma {\dfrac 1 2 \paren {a + b + 1} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } \map \Gamma {c - \dfrac a 2 + 1} \map \Gamma {\dfrac a 2} } {\map \Gamma c \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } \map \Gamma {\dfrac 1 2 \paren {1 + b} } }
      | c = After substitution
}}
{{eqn | r = \dfrac {2^{2 c - 1} \map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 + c} \map \Gamma {\dfrac 1 2 \paren {1 + a + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {2^{a - 1} 2^{2 c - a} \map \Gamma {\dfrac 1 2 \paren {1 + a} } \map \Gamma {\dfrac 1 2 \paren {1 + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } }
      | c = simplifying: $\map \Gamma c$, $\map \Gamma {c - \dfrac a 2 + 1}$, $\map \Gamma {\dfrac 1 2}$ and $\map \Gamma {\dfrac a 2}$ cancel
}}
{{eqn | r = \dfrac {\map \Gamma {\dfrac 1 2} \map \Gamma {\dfrac 1 2 + c} \map \Gamma {\dfrac 1 2 \paren {1 + a + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a - b} + c } } {\map \Gamma {\dfrac 1 2 \paren {1 + a} } \map \Gamma {\dfrac 1 2 \paren {1 + b} } \map \Gamma {\dfrac 1 2 \paren {1 - a} + c } \map \Gamma {\dfrac 1 2 \paren {1 - b} + c } }
      | c = simplifying further: Product of Powers, Quotient of Powers
}}
{{end-eqn}}
{{qed}}