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Proof step 14 from proof in-mkbag-del-two (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 15:
m0 - (m0 = 0 ⊃ 0;1) = 0, A[n0] = A[n], A[m] = A[n], count(A[n], mkbag(A, i)) = m0 + 1, m < i, n < i, n0 < i, m < # A, n < # A, n0 < # A, i ≤ # A ⊦ count(A[n], mkbag(A, i)) = 0, n = n0, n = m, n0 = m

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [count(A[n], mkbag(A, i)), m0]

Additional information

Used simplifier rules: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: count(A[n], mkbag(A, i)), m0