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Proof step 4 from proof gc-heap-ididx-h (previous step)

id-idx(lpt0, lpt1), m0 +1 ≤ # lpt1, ∀ lpt0, lpt1, nx. id-idx(lpt0, lpt1) ∧ m0 ≤ # lpt1 → gc-heap(lpt1, nx, m0) = gc-heap(lpt0, nx, m0) ⊦ gc-heap(lpt1, nx, m0 +1) = gc-heap(lpt0, nx, m0 +1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m0 < # lpt1, id-idx(lpt0, lpt1), ∀ lpt0, lpt1, nx. id-idx(lpt0, lpt1) ∧ m0 ≤ # lpt1 → gc-heap(lpt1, nx, m0) = gc-heap(lpt0, nx, m0) ⊦ gc-heap(lpt1, nx, m0) ∪ (¬ m0 ∈ nx ∧ lpt1[m0].flags = LP_GROUP_NODES ⊃ ∅ ++ key-index(lpt1[m0].ref-size, m0);∅) = gc-heap(lpt0, nx, m0 + 1)

Rule argument

Additional information

Used simplifier rules: gc-heap-rec: ⊦ n < # lpt → gc-heap(lpt, nx, n + 1) = gc-heap(lpt, nx, n) ∪ (¬ n ∈ nx ∧ lpt[n].flags = LP_GROUP_NODES ⊃ {key-index(lpt[n].ref-size, n)} ;∅ );% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ {ki} = ∅ ++ ki; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m +1 = m + 1; (s)