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Proof step 5 from proof abs-log-no-change-eq (previous step)

∀ gstore0, gstore1. (∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist)) → abs-log(nx0, gstore0) = abs-log(nx0, gstore1), ∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist) ⊦ gstore0[lnum].ladrlist.to-adr(lnum) + abs-log(nx0, gstore0) = gstore1[lnum].ladrlist.to-adr(lnum) + abs-log(nx0, gstore1)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
gstore0[lnum] = groupblock(gndx, lax, m), ∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist), ∀ gstore0, gstore1. (∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist)) → abs-log(nx0, gstore0) = abs-log(nx0, gstore1) ⊦ lax.to-adr(lnum) + abs-log(nx0, gstore0) = gstore1[lnum].ladrlist.to-adr(lnum) + abs-log(nx0, gstore1)

Rule argument

spec: ablock, inst: group-store name: elim seq: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m) subst: [gblock, gndx, lax, m] → [gstore0[lnum], gndx, lax, m]

Additional information

Used simplifier rules: ⊦ (gndx = gblock.gndlist ∧ lax = gblock.ladrlist) ∧ m = gblock.ref-size ↔ gblock = groupblock(gndx, lax, m); (elim) Used terms: gstore0[lnum], gndx, lax, m