Commutative Property Variation 1
Substitution 2
Substitution 6
Zero Plus a
Addi to Zero
Transitive Property of Equality Variation 2
Addi Zero Insn 0
Transitive Property Application 2
Pc 0
Addi Zero Insn 1
Pc 1
Byte 3 Stays the Same 1
Addi Zero Insn 2
Pc 2
Byte 3 Stays the Same 2
Byte 4 Stays the Same 2
Addi Insn 3
Pc 3
Byte 3 Stays the Same 3
Byte 4 Stays the Same 3
Addi Insn 4
Pc 4
Byte 3 Stays the Same 4
Byte 5 Stays the Same 4
Truth Propagation Property 2
Beq No Branch 5
Byte 3 Stays the Same 5
Byte 4 Stays the Same 5
Byte 5 Stays the Same 5
Jump Help 6
Byte 3 Stays the Same 6
Byte 4 Stays the Same 6
Byte 5 Stays the Same 6
Addi Insn 7
Pc 7
Byte 3 Stays the Same 7
Byte 4 Stays the Same 7
Addi Insn 8
Pc 8
Byte 3 Stays the Same 8
Byte 5 Stays the Same 8
Beq No Branch 9
Byte 3 Stays the Same 9
Byte 4 Stays the Same 9
Byte 5 Stays the Same 9
Jump Help 10
Byte 3 Stays the Same 10
Byte 4 Stays the Same 10
Byte 5 Stays the Same 10
Addi Insn 11
Pc 11
Byte 3 Stays the Same 11
Byte 4 Stays the Same 11
Addi Insn 12
Pc 12
Byte 3 Stays the Same 12
Byte 5 Stays the Same 12
Transitive Property of Equality Variation 1
Beq Branch
Byte 3 Stays the Same 13
Byte 4 Stays the Same 13
Byte 5 Stays the Same 13
Loop

Proof: Pc 1

Let's prove the following theorem:

if the following are true:
  • instruction #1 is addi dst=4 src=0 imm=0
  • the PC at time 1 = 1

then the PC at time 2 = 2

Proof:

View as a tree | View dependent proofs | Try proving it

Given
1 instruction #1 is addi dst=4 src=0 imm=0
2 the PC at time 1 = 1
Proof Table
# Claim Reason
1 the PC at time (1 + 1) = 1 + 1 if instruction #1 is addi dst=4 src=0 imm=0 and the PC at time 1 = 1, then the PC at time (1 + 1) = 1 + 1
2 1 + 1 = 2 1 + 1 = 2
3 the PC at time (1 + 1) = the PC at time 2 if 1 + 1 = 2, then the PC at time (1 + 1) = the PC at time 2
4 the PC at time 2 = 2 if the PC at time (1 + 1) = 1 + 1 and the PC at time (1 + 1) = the PC at time 2 and 1 + 1 = 2, then the PC at time 2 = 2
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