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Proof step 5 from proof subset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ st[a] = d ∧ a ∈ st → st[a, d] = st
Proof step 7:
¬(st1[a] = d ∧ a ∈ st1), # st0 < # st1, a ∈ st1, st1[a, d] ⊆ st0, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1[a, d] = ∅, st1 = ∅
Proof step 15:
st1[a] = d ∧ a ∈ st1, # st0 < # st1, a ∈ st1, st1 ⊆ st0, ∀ st0. # st0 < # st1 → ¬ st1 ⊆ st0 ⊦ st1 = ∅, st1 = ∅

Rule argument

spec: store inst: name: ins seq: ⊦ st[a] = d ∧ a ∈ st → st[a, d] = st subst: [d, a, st] → [d, a, st1] flags: rewrite, current

Additional information

Used terms: d, a, st1