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Proof step 25 from proof in-mkbag-del-two (previous step)

count(A[n0], mkbag(A, i)) - (count(A[n0], mkbag(A, i)) = 0 ⊃ 0;1) = 0, A[m] = A[n0], m < i, n < i, n0 < i, m < # A, n < # A, n0 < # A, i ≤ # A ⊦ A[n0] = A[n], n0 = m

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 26:
count(A[n0], mkbag(A, i)) - 0 = 0, A[m] = A[n0], count(A[n0], mkbag(A, i)) = 0, n0 < # A, n < # A, m < # A, n0 < i, n < i, m < i, i ≤ # A ⊦ A[n0] = A[n], n0 = m
Proof step 29:
count(A[n0], mkbag(A, i)) - 1 = 0, A[m] = A[n0], n0 < # A, n < # A, m < # A, n0 < i, n < i, m < i, i ≤ # A ⊦ A[n0] = A[n], count(A[n0], mkbag(A, i)) = 0, n0 = m

Rule argument

pos: left 1

Proof step 25 from proof in-mkbag-del-two (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information