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Proof step 10 from proof gc-heap-ididx-h (previous step)

lpt1[m0] = mklp(m, m1, lpf, m2), lpt0[m0] = mklp(m3, m4, lpf0, m5), m0 < # lpt0, 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 ∧ lpf = LP_GROUP_NODES ⊃ ∅ ++ key-index(m, m0);∅) = gc-heap(lpt0, nx, m0) ∪ (¬ m0 ∈ nx ∧ lpf0 = LP_GROUP_NODES ⊃ ∅ ++ key-index(m3, m0);∅)

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: lpt0, lpt1, nx abort: F discard: T computed: F

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)%id-idx-fwd-0: ⊦ lpt0[n] = mklp(n0, n1, lpf0, n2) ∧ lpt1[n] = mklp(m0, m1, lpf1, m2) ∧ n < # lpt0 ∧ id-idx(lpt0, lpt1) → n0 = m0 ∧ n1 = m1 ∧ lpf0 = lpf1; (forward) from specification gc-heap-rename Used terms: lpt0, lpt1, nx