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Proof step 5 from proof perm-shift-one (previous step)

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

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 6:
(mkbag(A, # A) -- A[m + 1]) ++ a = mkbag(A0, # A0), m + 1 < # A, m < # A ⊦ (mkbag(A, # A) -- A[m] -- A[m + 1]) ++ a ++ A[m] = (mkbag(A, # A) -- A[m + 1]) ++ a, a = A[m + 1]

Rule argument

pos: right 1

Additional information

Used simplifier rules: mkbag-del-add-content: m < i, i ≤ # A ⊦ (mkbag(A, i) -- A[m]) ++ A[m] = mkbag(A, i);% ⊦ m ≤ m; (s)