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Proof step 3 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)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
∀ 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)

Rule argument

indvar: nx

Additional information

Used simplifier rules: abs-log-base: ⊦ abs-log([], gstore) = [];%nat-list freely generated by [], +;;