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Proof step 13 from proof And-forall (previous step)

adP(a, d)(H), adP0(a, d)(H), adP0(a0, st0[a0])(H0), a0 ∈ st0, ((adP0 ∧_ st0))(H), ∀ 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(a0, st0[a0])(H0), a ∈ st0, ((adP ∧_ st0))(H)

User applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a0, H0 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, H0