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Proof step 8 from proof keys-same-fs (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 9:
fs0 | (ax.toSet ++ adr) = fs1 | (ax.toSet ++ adr), fs0 | ax.toSet = fs1 | ax.toSet → keys(fs0, ax) = keys(fs1, ax) ⊦ fs0[adr].key = fs1[adr].key ∧ keys(fs0, ax) = keys(fs1, ax)

Rule argument

pos: left 2 terms: fs0, fs1 abort: F discard: T computed: T

Additional information

Used terms: fs0, fs1