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Proof step 15 from proof difference (previous step)

count(a0, (bag0 -- a0) \ bag) = count(a0, bag0) - 1 - count(a0, bag), count(a0, bag) + 1 < count(a0, bag0) ⊦ count(a0, bag0) = 0, count(a0, bag0) - 1 - count(a0, bag) = count(a0, bag0) - count(a0, bag) + 1

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat inst: name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [count(a0, bag0), m] flags:

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: count(a0, bag0), m