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Proof step 7 from proof precise-mapex (previous step)

∀ r0. (r0 ∈ (∅ ++ r)[r, o] ∪ H0 ↔ r0 ∈ (∅ ++ r)[r, o0] ∪ H) ∧ (r0 ∈ (∅ ++ r)[r, o] ∪ H0 → ((∅ ++ r)[r, o] ∪ H0)[r0] = ((∅ ++ r)[r, o0] ∪ H)[r0]) ⊦ (∅ ++ r)[r, o] = (∅ ++ r)[r, o0], r = null, r ∈ H0, r ∈ H

User applied rule all left

And obtained the following new goals

Proof step 8:
r ∈ (∅ ++ r)[r, o] ∪ H0 ↔ r ∈ (∅ ++ r)[r, o0] ∪ H, r ∈ (∅ ++ r)[r, o] ∪ H0 → ((∅ ++ r)[r, o] ∪ H0)[r] = ((∅ ++ r)[r, o0] ∪ H)[r] ⊦ (∅ ++ r)[r, o] = (∅ ++ r)[r, o0], r = null, r ∈ H, r ∈ H0

Rule argument

pos: left 1 terms: r abort: F discard: T computed: F

Additional information

Used terms: r