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Proof step 13 from proof subset-size (previous step)

a ∈ st0, st1 -- a ⊆ st0 -- a, a ∈ st1, # st0 < # st1, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a] = st0[a], st1 = ∅, st1[a, st0[a]] = ∅, st1 ⊆ st0[a, st1[a]]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ st ⊆ st0[a, d] ↔ st ⊆ st0 ∧ ¬ a ∈ st ∨ a × d ∈ st ∧ st -- a ⊆ st0 -- a

Rule argument

spec: store inst: name: subset-put-01 seq: ⊦ st ⊆ st0[a, d] ↔ st ⊆ st0 ∧ ¬ a ∈ st ∨ a × d ∈ st ∧ st -- a ⊆ st0 -- a subst: [d, a, st0, st] → [st1[a], a, st0, st1] flags: rewrite, current

Additional information

Used simplifier rules: subset-elem-eq-01: ⊦ b ∈ st ∧ st[b] ≠ st1[b] → ¬ st ⊆ st1;%in-pair: ⊦ a × d ∈ st ↔ ¬ ¬(a ∈ st ∧ st[a] = d); Used terms: st1[a], a, st0, st1