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Proof step 24 from proof perm-shift-two (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 25:
(A[m + 1] = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ A[m + 1]) ++ b = mkbag(A0, # A0), A[m + 1] = A[m + 2], m + 1 < # A, m + 2 < # A, m < # A ⊦ (mkbag(A, # A) -- A[m] -- A[m + 2]) ++ A[m] = mkbag(A, # A) -- A[m + 1], b = A[m + 2]
Proof step 28:
(A[m + 1] = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ A[m + 1]) ++ b = mkbag(A0, # A0), m + 1 < # A, m + 2 < # A, m < # A ⊦ (mkbag(A, # A) -- A[m] -- A[m + 2]) ++ A[m] = (mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ A[m + 1], A[m + 1] = A[m + 2], b = A[m + 2]

Rule argument

pos: right 1

Proof step 24 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