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Proof step 21 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 + 1 < # A, m + 2 < # A, m < # A ⊦ ((a = A[m + 1] ⊃ mkbag(A, # A) -- A[m];(mkbag(A, # A) -- A[m] -- A[m + 1]) ++ a) -- A[m + 2]) ++ b ++ A[m] = (a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a) ++ b, b = A[m + 2]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
(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 + 1] ⊃ mkbag(A, # A) -- A[m];(mkbag(A, # A) -- A[m] -- A[m + 1]) ++ a) -- A[m + 2]) ++ A[m] = (a = A[m + 2] ⊃ mkbag(A, # A) -- A[m + 1];(mkbag(A, # A) -- A[m + 1] -- A[m + 2]) ++ a), b = A[m + 2]

Rule argument

Additional information

Used simplifier rules: ⊦ bag0 ++ a ++ b = bag1 ++ a ↔ bag0 ++ b = bag1; (ss)