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Proof step 10 from proof id-idx-write-both (previous step)

# lpt1 = # lpt0, lpt0[n, mklp(n0, n1, lpf, n2)][n3] = mklp(m, m0, lpf0, m1), n3 < # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ m = lpt1[n, mklp(n0, n1, lpf, m2)][n3].ref-size ∧ m0 = lpt1[n, mklp(n0, n1, lpf, m2)][n3].size ∧ lpf0 = lpt1[n, mklp(n0, n1, lpf, m2)][n3].flags

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 11:
# lpt1 = # lpt0, lpt0[n, mklp(n0, n1, lpf, n2)][n3] = mklp(m, m0, lpf0, m1), lpt1[n, mklp(n0, n1, lpf, m2)][n3] = mklp(m3, m4, lpf1, m5), n3 < # lpt0, ∀ n. n < # lpt0 → lpt0[n].ref-size = lpt1[n].ref-size ∧ lpt0[n].size = lpt1[n].size ∧ lpt0[n].flags = lpt1[n].flags ⊦ m = m3 ∧ m0 = m4 ∧ lpf0 = lpf1

Rule argument

spec: lprops name: elim seq: ⊦ ((m = lp.ref-size ∧ m0 = lp.size) ∧ lpf = lp.flags) ∧ m1 = lp.gcheapidx ↔ lp = mklp(m, m0, lpf, m1) subst: [lp, m, m0, lpf, m1] → [lpt1[n, mklp(n0, n1, lpf, m2)][n3], m3, m4, lpf1, m5]

Additional information

Used simplifier rules: ⊦ ((m = lp.ref-size ∧ m0 = lp.size) ∧ lpf = lp.flags) ∧ m1 = lp.gcheapidx ↔ lp = mklp(m, m0, lpf, m1); (elim) Used terms: lpt1[n, mklp(n0, n1, lpf, m2)][n3], m3, m4, lpf1, m5