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Proof step 7 from proof payload-copy-right-upper (previous step)

n + nused0 + 1 < # zbrar, n + 1 < # zbrar, ∀ n, zbrar. n + nused0 < # zbrar → payload(copy(zbrar, n, zbrar, n + 1, nused0), n + 1, nused0) = payload(zbrar, n, nused0) ⊦ (zbrar[2 * n - n].entry? ⊃ payload(copy(zbrar, n, zbrar, n + 1, nused0 + 1), 2 + n, nused0)[zbrar[2 * n - n].key, zbrar[2 * n - n].adr] ;payload(copy(zbrar, n, zbrar, n + 1, nused0 + 1), 2 + n, nused0) ) = payload(zbrar, n, nused0 + 1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
2 * n = n0 + n, n + 1 < # zbrar, n + nused0 + 1 < # zbrar, n ≤ 2 * n, ∀ n, zbrar. n + nused0 < # zbrar → payload(copy(zbrar, n, zbrar, n + 1, nused0), n + 1, nused0) = payload(zbrar, n, nused0) ⊦ (zbrar[n0].entry? ⊃ payload(copy(zbrar, n, zbrar, n + 1, nused0 + 1), 2 + n, nused0)[zbrar[n0].key, zbrar[n0].adr] ;payload(copy(zbrar, n, zbrar, n + 1, nused0 + 1), 2 + n, nused0) ) = payload(zbrar, n, nused0 + 1)

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [2 * n, n, n0]

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ ¬ 2 * n < n; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: 2 * n, n, n0