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Proof step 3 from proof Diff-rec-rev (previous step)

∀ st, a. (st \ s0) -- a = (st -- a) \ s0 ⊦ (st \ (s0 ++ a0)) -- a = (st -- a) \ (s0 ++ a0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ st, a. (st \ s0) -- a = (st -- a) \ s0 ⊦ (st \ s0) -- a0 -- a = ((st -- a) \ s0) -- a0

Rule argument

Additional information

Used simplifier rules: Diff-rec: ⊦ st \ (s ++ a) = (st \ s) -- a;