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Proof step 35 from proof perm-shift-two (previous step)

(a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a) ++ b = mkbag(A0, # A0), m < # A, m + 2 < # A, m + 1 < # A ⊦ (a = A[m + 2] ⊃ mkbag(A, # A) -- A[m] -- A[m + 1];(mkbag(A, # A) -- A[m] -- A[m + 1] -- A[m + 2]) ++ a) ++ A[m] = (a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a), a = A[m + 1], b = A[m + 2]

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 36:
(a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a) ++ b = mkbag(A0, # A0), a = A[m + 2], m + 1 < # A, m + 2 < # A, m < # A ⊦ (mkbag(A, # A) -- A[m] -- A[m + 1]) ++ A[m] = (a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a), a = A[m + 1], b = A[m + 2]
Proof step 39:
(a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a) ++ b = mkbag(A0, # A0), m + 1 < # A, m + 2 < # A, m < # A ⊦ (mkbag(A, # A) -- A[m] -- A[m + 1] -- A[m + 2]) ++ a ++ A[m] = (a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a), a = A[m + 1], a = A[m + 2], b = A[m + 2]

Rule argument

pos: right 1

Proof step 35 from proof perm-shift-two (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information