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

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

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_GROUP_NODES ↔ n ∈ gstore) ∧ (n ∈ gstore ∧ ¬ n ∈ ginval → abs-leb(avol[n], gstore[n], lpt1[n])) ∧ (n ∈ gstore → lpt1[n].ref-size = gstore[n].ref-size)) ↔ (∀ n. (n < # lpt0 ∧ lpt0[n].flags = LP_GROUP_NODES ↔ n ∈ gstore) ∧ (n ∈ gstore ∧ ¬ n ∈ ginval → abs-leb(avol[n], gstore[n], lpt0[n])) ∧ (n ∈ gstore → lpt0[n].ref-size = gstore[n].ref-size))

Rule argument

Additional information

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