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Proof step 1 from proof remove-refsize0-one (conclusion)

⊦ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ st1 = st2 ↔ (∀ n. (n ∈ st1 ↔ n ∈ st2) ∧ (n ∈ st1 → st1[n] = st2[n]))
Proof step 3:
⊦ ∀ n0. (n0 ∈ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) ↔ n0 ∈ (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])) ∧ ( n0 ∈ remove-refsize0((∅)[n, mkblock(x, lax, m0)]) → remove-refsize0((∅)[n, mkblock(x, lax, m0)])[n0] = (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)])[n0])

Rule argument

spec: store, inst: ablock-store-nat-store name: k-Extension seq: ⊦ st1 = st2 ↔ (∀ n. (n ∈ st1 ↔ n ∈ st2) ∧ (n ∈ st1 → st1[n] = st2[n])) subst: [st2, st1] → [(m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)]), remove-refsize0((∅)[n, mkblock(x, lax, m0)])] flags: <all paths>

Additional information

Used terms: (m0 = 0 ⊃ ∅;(∅)[n, mkblock(x, lax, m0)]), remove-refsize0((∅)[n, mkblock(x, lax, m0)])