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Proof step 4 from proof mkbag-update (previous step)

m < # A, m ≤ n0, ∀ m, A, a. m < n0 ∧ m < # A → mkbag(A[m, a], n0) = (mkbag(A, n0) -- A[m]) ++ a ⊦ mkbag(A[m, a], n0) ++ A[m, a][n0] = (A[n0] = A[m] ⊃ mkbag(A, n0);(mkbag(A, n0) -- A[m]) ++ A[n0]) ++ a

User applied rule case distinction

And obtained the following new goals

Proof step 5:
A[n0] = A[m], m < # A, m ≤ n0, ∀ m, A, a. m < n0 ∧ m < # A → mkbag(A[m, a], n0) = (mkbag(A, n0) -- A[m]) ++ a ⊦ mkbag(A[m, a], n0) ++ A[m, a][n0] = mkbag(A, n0) ++ a
Proof step 13:
m < # A, m ≤ n0, ∀ m, A, a. m < n0 ∧ m < # A → mkbag(A[m, a], n0) = (mkbag(A, n0) -- A[m]) ++ a ⊦ mkbag(A[m, a], n0) ++ A[m, a][n0] = (mkbag(A, n0) -- A[m]) ++ A[n0] ++ a, A[n0] = A[m]

Rule argument

pos: right 1

Proof step 4 from proof mkbag-update (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information