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Proof step 2 from proof abs-log-no-change-eq (previous step)

∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist) ⊦ abs-log(nx, gstore0) = abs-log(nx, gstore1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ n. (n ∈ gstore0 ↔ n ∈ gstore1) ∧ (n ∈ gstore0 → gstore1[n].gndlist = gstore0[n].gndlist ∧ gstore1[n].ladrlist = gstore0[n].ladrlist) ⊦ abs-log(nx, gstore0) = abs-log(nx, gstore1)

Proof step 2 from proof abs-log-no-change-eq (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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