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Proof step 28 from proof max-rem-notsame (previous step)

m < bag0.max, bag0 = ∅ ++ bag0.max ∨ bag0.max = (bag0 -- bag0.max).max ⊦ bag0.max = (bag0 = ∅ ++ bag0.max ⊃ m;(m < (bag0 -- bag0.max).max ⊃ (bag0 -- bag0.max).max;m)), bag0 = ∅, count(bag0.max, bag0) = 1, m = bag0.max

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 29:
bag0 = ∅ ++ bag0.max, m < bag0.max, bag0 = ∅ ++ bag0.max ∨ bag0.max = (bag0 -- bag0.max).max ⊦ bag0.max = m, bag0 = ∅, count(bag0.max, bag0) = 1

Rule argument

pos: right 1

Additional information

Used simplifier rules: lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic