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Proof step 5 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 6:
⊦ a ≠ b → (st -- b)[a] = st[a]
Proof step 7:
∀ st, a. (st \ s0)[a] = st[a] ∨ a ∈ s0 ⊦ (st \ s0)[a] = st[a], a = a0, a ∈ s0 ++ a0

Rule argument

spec: store name: Del-at seq: ⊦ a ≠ b → (st -- b)[a] = st[a] subst: [a, b, st] → [a, a0, st \ s0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ (st -- a)[a] = (∅)[a]; (s)% ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (ws) Used terms: a, a0, st \ s0