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Proof step 6 from proof bag-add-rev (previous step)

⊦ (n0 = m ⊃ 1;0) + count(n0, bag ++ n) = count(n0, bag ++ m ++ n)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
n0 = m ⊦ 1 + count(n0, bag ++ n) = count(n0, bag ++ m ++ n)
Proof step 11:
⊦ 0 + count(n0, bag ++ n) = count(n0, bag ++ m ++ n), n0 = m

Rule argument

pos: right 1

Additional information