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Proof step 7 from proof Diff-del-rw (previous step)

⊦ (a0 ∈ st -- a ∧ ¬ a0 ∈ s ↔ a0 ∈ (st \ s) -- a) ∧ (a0 ∈ st -- a ∧ ¬ a0 ∈ s → (st -- a)[a0] = ((st \ s) -- a)[a0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st
Proof step 9:
⊦ (a0 ∈ st -- a ∧ ¬ a0 ∈ s ↔ a0 ≠ a ∧ a0 ∈ st \ s) ∧ (a0 ∈ st -- a ∧ ¬ a0 ∈ s → (st -- a)[a0] = ((st \ s) -- a)[a0])

Rule argument

spec: store name: Del-in seq: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st subst: [b, st, a] → [a, st \ s, a0] flags: <all paths>

Additional information

Used terms: a, st \ s, a0