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Proof step 4 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), ∀ 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) ⊦ abs-log(lnum + nx0, gstore0) = abs-log(lnum + nx0, gstore1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ 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)

Rule argument

Additional information

Used simplifier rules: abs-log-rec: ⊦ abs-log(n ' + nx, gstore) = gstore[n].ladrlist.to-adr(n) + abs-log(nx, gstore);%a: ⊦ (nx + nx0) + nx1 = nx + nx0 + nx1; from specification list[nat-list-set]% ⊦ n + nx = n ' + nx; (s)