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Proof step 9 from proof oplus-shift-put-l (previous step)

⊦ a0 = a, (a0 ∈ st0[a, d] ⊕ st1 ↔ a0 ∈ st0 ⊕ st1) ∧ (a0 ∈ st0[a, d] ⊕ st1 → (st0[a, d] ⊕ st1)[a0] = (st0 ⊕ st1)[a0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ a ∈ st0 → (st0 ⊕ st1)[a] = st0[a]
Proof step 11:
¬ a0 ∈ st0[a, d] ⊦ a0 = a, (a0 ∈ st0[a, d] ⊕ st1 ↔ a0 ∈ st0 ⊕ st1) ∧ (a0 ∈ st0[a, d] ⊕ st1 → (st0[a, d] ⊕ st1)[a0] = (st0 ⊕ st1)[a0])

Rule argument

spec: inst: name: oplus-l-at-left seq: ⊦ a ∈ st0 → (st0 ⊕ st1)[a] = st0[a] subst: [a, st1, st0] → [a0, st1, st0[a, d]] flags: rewrite, current

Additional information

Used simplifier rules: oplus-l-at-left: ⊦ a ∈ st0 → (st0 ⊕ st1)[a] = st0[a];%oplus-l-in: ⊦ a ∈ st0 ⊕ st1 ↔ a ∈ st0 ∨ a ∈ st1;%in-neq: ⊦ a ≠ b → (a ∈ st[b, d] ↔ a ∈ st); from specification store%At-other: ⊦ a ≠ b → st[b, d][a] = st[a]; from specification store Used terms: a0, st1, st0[a, d]