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Proof step 5 from proof And-discard (previous step)

∀ s0, P. # s0 < # s → true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P ⊦ # s = 0, s = ∅, (λ a. P) ∧_ s = P

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2)
Proof step 7:
∀ s0, P. # s0 < # s → true → # s0 = 0 ∨ s0 = ∅ ∨ (λ a. P) ∧_ s0 = P ⊦ # s = 0, (λ a. P) ∧_ s = P, ∀ a. a ∈ s ↔ a ∈ ∅

Rule argument

spec: set-basic name: Extension seq: ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2) subst: [s2, s1] → [∅, s] flags: <all paths>

Additional information

Used terms: ∅, s