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Proof step 5 from proof dom-restrict (previous step)

s ⊆ dom st ⊦ a ∈ dom st \ dom st \ s ↔ a ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2

Rule argument

spec: set-union inst: name: Difference seq: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2 subst: [s2, s1, a] → [dom st \ s, dom st, a] flags: rewrite

Additional information

Used simplifier rules: Dom: ⊦ a ∈ dom st ↔ a ∈ st;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; from specification set-union%sub-in: ⊦ a ∈ s ∧ s ⊆ s0 → a ∈ s0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic Used terms: dom st \ s, dom st, a