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

aP0(a)(H), ((aP0 ∧_ s0))(H), ∀ a0, H0. (a0 = a ∨ a0 ∈ s0) ∧ aP0(a0)(H0) → aP(a0)(H0), ∀ s1, aP0, H, aP. # s1 < #(s0 ++ a) → ((aP0 ∧_ s1))(H) ∧ (∀ a, H. a ∈ s1 ∧ aP0(a)(H) → aP(a)(H)) → ((aP ∧_ s1))(H) ⊦ a ∈ s0, aP(a)(H) ∧ ((aP ∧_ s0))(H)

User applied rule all left

And obtained the following new goals

Proof step 14:
aP0(a)(H), ((aP0 ∧_ s0))(H), (a = a ∨ a ∈ s0) ∧ aP0(a)(H) → aP(a)(H), ∀ a0, H0. (a0 = a ∨ a0 ∈ s0) ∧ aP0(a0)(H0) → aP(a0)(H0), ∀ s1, aP0, H, aP. # s1 < #(s0 ++ a) → ((aP0 ∧_ s1))(H) ∧ (∀ a, H. a ∈ s1 ∧ aP0(a)(H) → aP(a)(H)) → ((aP ∧_ s1))(H) ⊦ a ∈ s0, aP(a)(H) ∧ ((aP ∧_ s0))(H)

Rule argument

pos: left 3 terms: a, H abort: F discard: F computed: T

Additional information

Used terms: a, H