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Proof step 9 from proof map-putex (previous step)

((r => o1 * P))((H0 ++ r)[r, o1]) ⊦ r = null, r ∈ H0, P((H0 ++ r)[r, o] -- r) ∧ r ∈ (H0 ++ r)[r, o] ∧ (H0 ++ r)[r, o][r] = o

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
((r => o1 * P))((H0 ++ r)[r, o1]) ⊦ r = null, P(H0), r ∈ H0

Rule argument

Additional information

Used simplifier rules: ⊦ H[r0, o][r] = (r = r0 ∧ r0 ∈ H ⊃ o;H[r]); (s)% ⊦ r ∈ H[r0, o] ↔ r ∈ H; (s)%¬ r ∈ H, r ≠ null ⊦ (H ++ r) -- r = H; (s)%r ∈ H ⊦ H[r, o] -- r = H -- r; (s)%r0 ≠ null, ¬ r0 ∈ H ⊦ r ∈ H ++ r0 ↔ r = r0 ∨ r ∈ H; (s)