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Proof step 10 from proof keys-same-fs (previous step)

fs0 | (ax.toSet ++ adr) = fs1 | (ax.toSet ++ adr), fs0 | ax.toSet = fs1 | ax.toSet, fs0 | ax.toSet = fs1 | ax.toSet → keys(fs0, ax) = keys(fs1, ax) ⊦ fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
keys(fs0, ax) = keys(fs1, ax), fs0 | ax.toSet = fs1 | ax.toSet, fs0 | (ax.toSet ++ adr) = fs1 | (ax.toSet ++ adr) ⊦ fs0[adr].key = fs1[adr].key

Rule argument

Additional information