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Proof step 3 from proof Diff-get (previous step)

∀ st, a. (st \ s0)[a] = st[a] ∨ a ∈ s0 ⊦ (st \ (s0 ++ a0))[a] = st[a], a ∈ s0 ++ a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ st \ (s ++ a) = (st \ s) -- a
Proof step 5:
∀ st, a. (st \ s0)[a] = st[a] ∨ a ∈ s0 ⊦ ((st \ s0) -- a0)[a] = st[a], a ∈ s0 ++ a0

Rule argument

spec: <toplevel> name: Diff-rec seq: ⊦ st \ (s ++ a) = (st \ s) -- a subst: [a, s, st] → [a0, s0, st] flags: <all paths>

Additional information

Used terms: a0, s0, st