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Proof step 2 from proof id-idx-abs-istore (previous step)

id-idx(lpt0, lpt1) ⊦ abs-istore(avol, lpt1, istore) ↔ abs-istore(avol, lpt0, istore)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
id-idx(lpt0, lpt1) ⊦ (∀ n. (n < # lpt1 ∧ lpt1[n].flags = LP_INDEX_NODES ↔ n ∈ istore) ∧ (n ∈ istore → abs-leb(avol[n], istore[n], lpt1[n]) ∧ lpt1[n].ref-size = istore[n].ref-size)) ↔ (∀ n. (n < # lpt0 ∧ lpt0[n].flags = LP_INDEX_NODES ↔ n ∈ istore) ∧ (n ∈ istore → abs-leb(avol[n], istore[n], lpt0[n]) ∧ lpt0[n].ref-size = istore[n].ref-size))

Rule argument

Additional information

Used simplifier rules: abs-istore-def: ⊦ abs-istore(avol, lpt, istore) ↔ (∀ n. (n < # lpt ∧ lpt[n].flags = LP_INDEX_NODES ↔ n ∈ istore) ∧ ( n ∈ istore → abs-leb(avol[n], istore[n], lpt[n]) ∧ lpt[n].ref-size = istore[n].ref-size));