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Proof step 11 from proof And-forall (previous step)

adP(a, d)(H), adP0(a, d)(H), ((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) ⊦ a ∈ st0, ((adP ∧_ st0))(H)

User applied rule apply induction

And obtained the following new goals

Proof step 12:
adP(a, d)(H), adP0(a, d)(H), ((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) ⊦ a ∈ st0, ((adP ∧_ st0))(H), ((adP0 ∧_ st0))(H) ∧ (∀ a, H. a ∈ st0 ∧ adP0(a, st0[a])(H) → adP(a, st0[a])(H))

Rule argument

subst: [st1, adP0, H, adP] → [st0, adP0, H, adP]

Additional information

Used simplifier rules: ⊦ ¬ a ∈ st → # st[a, d] = # st + 1; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ ¬ n < n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: st0, adP0, H, adP