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Proof step 8 from proof payload-copy-right-upper (previous step)

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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n0 + 1 < # zbrar, n0 + nused0 + 1 < # zbrar, ∀ 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, n0, zbrar, n0 + 1, nused0 + 1), 2 + n0, nused0)[zbrar[n0].key, zbrar[n0].adr] ;payload(copy(zbrar, n0, zbrar, n0 + 1, nused0 + 1), 2 + n0, nused0) ) = payload(zbrar, n0, nused0 + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ m ≤ 2 * m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%mult-c: ⊦ m * n = n * m; from specification nat-mult%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult% ⊦ m0 * 1 = m0; (s)%m2 ≠ 0 ⊦ (* m2) * m = m + n0 ↔ (m2 - 1) * m = n0; (s)