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Proof step 3 from proof keys-same-fs (previous step)

fs0 | ax1.toSet = fs1 | ax1.toSet ⊦ keys(fs0, ax1) = keys(fs1, ax1)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
fs0 | (adr + ax).toSet = fs1 | (adr + ax).toSet, ∀ fs0, fs1. fs0 | ax.toSet = fs1 | ax.toSet → keys(fs0, ax) = keys(fs1, ax) ⊦ keys(fs0, adr + ax) = keys(fs1, adr + ax)

Rule argument

indvar: ax1

Additional information

Used simplifier rules: address-list freely generated by [], +;;% ⊦ [].toSet = ∅; (s)% ⊦ fs | ∅ = ∅; (s)% ⊦ keys(fs, []) = []; (s)