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Proof step 8 from proof And-discard (previous step)

∀ st1, P. # st1 < # st0[a, d] → true → st1 = ∅ ∨ (λ a, d. P) ∧_ st1 = P ⊦ (P ∧ (λ a, d. P) ∧_ st0) = P, a ∈ st0

User applied rule apply induction

And obtained the following new goals

Proof step 9:
st0 = ∅ ∨ (λ a, d. P) ∧_ st0 = P, ∀ st1, P. # st1 < # st0[a, d] → true → st1 = ∅ ∨ (λ a, d. P) ∧_ st1 = P ⊦ (P ∧ (λ a, d. P) ∧_ st0) = P, a ∈ st0

Rule argument

subst: [st1, P] → [st0, P]

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, P