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

∀ a, st. a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0 ⊦ a ∈ (st \ s0) -- a0 ↔ a ∈ st ∧ a ≠ a0 ∧ ¬ a ∈ s0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st
Proof step 5:
∀ a, st. a ∈ st \ s0 ↔ a ∈ st ∧ ¬ a ∈ s0 ⊦ a ≠ a0 ∧ a ∈ st \ s0 ↔ a ∈ st ∧ a ≠ a0 ∧ ¬ a ∈ s0

Rule argument

spec: store inst: name: Del-in seq: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st subst: [b, st, a] → [a0, st \ s0, a] flags: rewrite

Additional information

Used terms: a0, st \ s0, a