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Proof step 17 from proof Sep-forall (previous step)

adP0(a, d)(H1), adP0(a0, st0[a0])(H), a0 ∈ st0, ((adP0 ⊗_ st0))(H0), H1 ⊥ H0, ∀ a0, H. a0 ∈ st0[a, d] ∧ adP0(a0, st0[a, d][a0])(H) → adP(a0, st0[a, d][a0])(H), ∀ st1, adP0, H, adP. # st1 < # st0[a, d] → ((adP0 ⊗_ st1))(H) ∧ (∀ a, H. a ∈ st1 ∧ adP0(a, st1[a])(H) → adP(a, st1[a])(H)) → ((adP ⊗_ st1))(H) ⊦ ((adP(a, d) * adP ⊗_ st0))(H1 ∪ H0), adP(a0, st0[a0])(H), a ∈ st0, adP(a, d)(H1) ∧ ((adP ⊗_ st0))(H0)

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a0, H abort: F discard: F computed: F

Additional information

Used simplifier rules: ⊦ b ∈ st ∧ ¬ a ∈ st → st[a, d][b] = st[b]; (s)% ⊦ a ∈ st[b, d] ↔ a = b ∨ a ∈ st; (ws) Used terms: a0, H