which, by algebraic manipulation, implies thatx=x0.
Surjective We are required to show that if f1 and f2 are surjective, then the composition of f1 and f2 is surjective as well.
∀i∈ A,∃j ∈ A: f1(j) =i ^
(A.40)
∀i∈ A,∃j ∈ A: f2(j) =i (A.41)
=⇒
∀i∈ A,∃j ∈ A: f2(f1(j)) =i. (A.42)
Letx00 be an instance ofi in Equation A.42; use this value fori in Equation A.41. Let x0 be an instance ofj in Equation A.41; use this value fori in Equation A.40. Finally, letx be an instance of j in Equation A.40; use this value forj in Equation A.42. The implication follows:
f1(x) = x0∧f2(x0) =x00 =⇒ f2(f1(x)) = x00
=⇒ f2(x0) =x00
=⇒ x00 =x00.
Permutes We are required to show
∀i∈ A: S(i) =S0(f1(i)) ^
(A.43)
∀i∈ A: S0(i) = S00(f2(i)) (A.44)
=⇒
∀i∈ A: S(i) =S00(f2(f1(i))). (A.45) Let x be the instantiation of i in Equation A.45. Use x for i in Equation A.43 and
f1(x) in Equation A.44. The implication follows:
S(x) = S0(f1(x))∧S0(f1(x)) = S00(f2(f1(x))) =⇒ S(x) =S00(f2(f1(x)))
=⇒ S(x) =S0(f1(x))
=⇒ S(x) =S(x).
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