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Proof step 25 from proof lst-in-hp (previous step)

rs1 ∩ rs0 = ∅, R[rs0, cf, af], ls(r0, x0)(H), ∀ r. r ∈ rs1 → af(r) = o, ∀ r1. (r1 ∈ rs1 ↔ r1 ∈ (∅ ++ r)[r, a × r0] ∪ H) ∧ (r1 ∈ rs1 → cf(r1) = ((∅ ++ r)[r, a × r0] ∪ H)[r1]) ⊦ r = null, r ∈ rs0, r ∈ rs1, r ∈ H

User applied rule all left

And obtained the following new goals

Proof step 26:
r ∈ rs1 ↔ r ∈ (∅ ++ r)[r, a × r0] ∪ H, rs1 ∩ rs0 = ∅, R[rs0, cf, af], ls(r0, x0)(H), r ∈ rs1 → cf(r) = ((∅ ++ r)[r, a × r0] ∪ H)[r], ∀ r. r ∈ rs1 → af(r) = o ⊦ r = null, r ∈ H, r ∈ rs1, r ∈ rs0

Rule argument

pos: left 5 terms: r abort: F discard: T computed: T

Additional information

Used terms: r