Assumed: N
  Assumed: h
  Proved: CongrH
  Set option: lean::pp::implicit
Theorem CongrH {a1 a2 b1 b2 : N} (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) :=
    Congr::explicit
        N
        (λ x : N, N)
        (h a1)
        (h b1)
        a2
        b2
        (Congr::explicit N (λ x : N, N → N) h h a1 b1 (Refl::explicit (N → N → N) h) H1)
        H2
Theorem CongrH::explicit (a1 a2 b1 b2 : N) (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) :=
    CongrH::explicit a1 a2 b1 b2 H1 H2
  Set option: lean::pp::implicit
Theorem CongrH {a1 a2 b1 b2 : N} (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) := Congr (Congr (Refl h) H1) H2
Theorem CongrH::explicit (a1 a2 b1 b2 : N) (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) := CongrH H1 H2
  Proved: Example1
  Set option: lean::pp::implicit
Theorem Example1 (a b c d : N) (H : a = b ∧ b = c ∨ a = d ∧ d = c) : (h a b) = (h c b) :=
    DisjCases::explicit
        (a = b ∧ b = c)
        (a = d ∧ d = c)
        ((h a b) = (h c b))
        H
        (λ H1 : a = b ∧ b = c,
           CongrH::explicit
               a
               b
               c
               b
               (Trans::explicit
                    N
                    a
                    b
                    c
                    (Conjunct1::explicit (a = b) (b = c) H1)
                    (Conjunct2::explicit (a = b) (b = c) H1))
               (Refl::explicit N b))
        (λ H1 : a = d ∧ d = c,
           CongrH::explicit
               a
               b
               c
               b
               (Trans::explicit
                    N
                    a
                    d
                    c
                    (Conjunct1::explicit (a = d) (d = c) H1)
                    (Conjunct2::explicit (a = d) (d = c) H1))
               (Refl::explicit N b))
  Proved: Example2
  Set option: lean::pp::implicit
Theorem Example2 (a b c d : N) (H : a = b ∧ b = c ∨ a = d ∧ d = c) : (h a b) = (h c b) :=
    DisjCases::explicit
        (a = b ∧ b = c)
        (a = d ∧ d = c)
        ((h a b) = (h c b))
        H
        (λ H1 : a = b ∧ b = c,
           CongrH::explicit
               a
               b
               c
               b
               (Trans::explicit
                    N
                    a
                    b
                    c
                    (Conjunct1::explicit (a = b) (b = c) H1)
                    (Conjunct2::explicit (a = b) (b = c) H1))
               (Refl::explicit N b))
        (λ H1 : a = d ∧ d = c,
           CongrH::explicit
               a
               b
               c
               b
               (Trans::explicit
                    N
                    a
                    d
                    c
                    (Conjunct1::explicit (a = d) (d = c) H1)
                    (Conjunct2::explicit (a = d) (d = c) H1))
               (Refl::explicit N b))
  Proved: Example3
  Set option: lean::pp::implicit
Theorem Example3 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c ∨ a = d ∧ d = c) : (h a b) = (h c b) :=
    DisjCases
        H
        (λ H1 : a = b ∧ b = e ∧ b = c, CongrH (Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1))) (Refl b))
        (λ H1 : a = d ∧ d = c, CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
  Proved: Example4
  Set option: lean::pp::implicit
Theorem Example4 (a b c d e : N) (H : a = b ∧ b = e ∧ b = c ∨ a = d ∧ d = c) : (h a c) = (h c a) :=
    DisjCases
        H
        (λ H1 : a = b ∧ b = e ∧ b = c,
           let AeqC := Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1)) in CongrH AeqC (Symm AeqC))
        (λ H1 : a = d ∧ d = c, let AeqC := Trans (Conjunct1 H1) (Conjunct2 H1) in CongrH AeqC (Symm AeqC))