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Proof step 6 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 ⊦ (b = A[m + 2] ⊃ mkbag(A[m, a], # A) -- A[m + 1];(mkbag(A[m, a], # A) -- A[m + 1] -- 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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
(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 ⊦ (b = A[m + 2] ⊃ (a = A[m + 1] ⊃ mkbag(A, # A) -- A[m];(mkbag(A, # A) -- A[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 + 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

Rule argument

Additional information

Used simplifier rules: mkbag-update: m < # A ⊦ m < n → mkbag(A[m, a], n) = (mkbag(A, n) -- A[m]) ++ a;% ⊦ (bag ++ a) -- b = (a = b ⊃ bag;(bag -- b) ++ a); (s)% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)